- The Latent Heat of Fusion (Lf) for water is 3.33 x 105 J/Kg. The Latent Heat of Vaporization (Lv) for water is 22.6 x 105 J/kg. Notice that it takes a lot more energy to turn water into steam than it does to melt ice. This makes sense - steam as a gas has a lot more kinetic energy per molecule than water - that is one reason it is so dangerous
- 500 g = 500/1,000 = 0.5 kg. specific latent heat of fusion = 334 kJ/kg (from the table above) 50 g of water was used. Calculate the latent heat of vaporisation of water
- The enthalpy of fusion of a substance, also known as (latent) heat of fusion is the change in its enthalpy resulting from providing energy, typically heat, to a specific quantity of the substance to change its state from a solid to a liquid, at constant pressure.For example, when melting 1 kg of ice (at 0 °C under a wide range of pressures), 333.55 kJ of energy is absorbed with no temperature.

Lf is the latent heat of fusion of water, 334 J/g * Latent heat is the heat absorbed or released as the result of a phase change*. This temperature change during a phase change is due to the energy released from the potential energy stored in the bonds between the particles, and not due to the energy release from the kinetic energy of the particle The Latent Heat of Fusion is the heat supplied to a solid body at the melting point when it changes state from solid to liquid. Latent heat of fusion for some metals are indicated below: 1 kJ/kg = 0.23884 cal/g = 0.23884 kcal/kg = 0.4299 Btu/lbm Latent heat of fusion is calculated a

The latent heat of fusion of a substance is the amount of heat required to convert a unit mass of the solid into liquid without change in temperature The latent heat of melting for some common solids are indicated below: 1 kJ/kg = 0.4299 Btu/lbm = 0.23884 kcal/k The Latent Heat Of Fusion Of Water Is 333000 J/kg. Change In Entropy: JAK This problem has been solved How much heat is required to convert 100.0 g of ice at -10.0°C to water at 50.0°C? the latent heat of fusion of water is 3.3x105 J/kg the latent heat of vaporization of water is 2.3x106 J/kg the specific heat of steam is 2.0 x 103 J/kg°C the specific heat of water is 4.2 x 103 J/kg°C the specific heat of ice is 2.1 x 103 J/kg° ** 0**.5kg of ice at -5degC is put into a vessel containing 2kg of water at 15deg C and mixed together, the result being a mixture of ice and water at** 0**degC.Calculate the final masses of ice and water, taking the water equivalent of the vessel as .15kg.The specific heat of ice is 2.04kJ/kg/K and the latent heat of fusion is 335kJ/kg In other words, if the value of the latent heat of fusion of water is 333.55 kJ / Kg, then the value of the latent heat of solidification or freezing of water will be -333.55 kJ / Kg. Latent heat of condensatio

For example, the latent heat of fusion of one kilogram of water, which is the amount of heat energy that must be supplied to convert 1 kg of ice without changing the temperature of the environment (which is kept at zero degrees celsius) is 333.55 kilojoules How much **heat** energy is required to change 0.5 kg of ice at 0°C into **water** at 25°C? [Specific **latent** **heat** **of** **fusion** **of** **water** = 334 000 **J/kg**; specific **heat** capacity of **water** = 4200 **J/**(kg K).] Answer: There are 2 processes involve when an ice is converted into **water** at 25°C. Ice at 0°C -----> **Water** at 0°C -----> **Water** at 25° ** Latent heat, heat of fusion, heat of vaporization - problems and solutions**. 1. Calculate the amount of heat added to 1 gram gold to change phase from solid to liquid. The heat of fusion for gold is 64.5 x 10 3 J/kg. Known : Mass (m) = 1 gram = 1 x 10-3 kg . Heat of fusion (L F) = 64.5 x 10 3 J/kg. Wanted : Heat (Q) Solution : Q = m L The latent heat of fusion is the energy required to melt 1 kg of a solid and has units of J kg −1. The latent heat of vaporisation is the energy required to evaporate 1 kg of a liquid and has the same units of J kg −1

Specific heat of ice = 2100J Kg−1K−1 Latent heat of fusion of ice = 3.36 ⋅ 105J kg−1 Specific heat capacity of water = 4200J kg−1K−1 Latent heat of vaporization of water = 2.25 ⋅ 106J kg− Question: 6. (3 Points). Determine The Latent Heat Of Fusion Of Mercury Using The Following Calorimeter Data: 2.00 Kg Of Solid Mercury At Its Melting Point Of - 39.0°C Is Placed In A 0.620 Kg Aluminum Calorimeter With 0.800 Kg Of Water At 12.80ºC; The Resulting Equilibrium Temperature Is 5.06°C Specific heat capacity of liquid water = 4200 J kg-1 K-1 Latent heat of fusion = 334 kJ kg-1. 2 MJ. In a steam turbine, 4.29 GJ of heat is used to turn 1200 kg of water at 20°C into steam. What is the temperature of the steam? Specific heat capacity of liquid water = 4200 J kg-1 K-

In other words 'The latent heat of fusion of a solid is the quantity of heat in joules required to convert 1 kilogram of the solid to liquid, with out any change in temperature. Ex: The latent heat of fusion of ice = 3.34 × 10 5 J/kg For a liquid to solidify at its freezing point, latent heat of fusion has to be removed from it measured in J/kg, NOTE: to fuse means to melt. In this experiment the latent heat of fusion of water will be determined by using the method of mixtures ΣQ=0, or QGained +QLost =0. Ice will be added to a calorimeter containing warm water. The heat energy lost by the water and calorimeter does tw Heat of fusion is the amount of heat energy required to change the state of matter of a substance from a solid to a liquid.It's also known as enthalpy of fusion. Its units are usually Joules per gram (J/g) or calories per gram (cal/g). This example problem demonstrates how to calculate the amount of energy required to melt a sample of water ice specific latent heat of vaporisation of water = 2.3 x 106J kg-1 (3 marks) ΔQ = ml Energy in one minute: 500 × 60 (1 mark) = m × 2.3 × 106 (1 mark) So m = 0.013 kg (1 mark) (c) Explain why, when she stops running, her temperature is likely to fall. She is not generating as much heat internally (1 mark) but i

- ed in five stages
- Some typical values for specific latent heat include: An input of 334,000 joules (J) of energy is needed to change 1 kg of ice into 1 kg of water at its melting point of 0°C. The same amount of..
- The energy is viewed as latent on the grounds that it is basically covered up inside the atoms until the stage change happens. The most well-known units of specific latent heat are joules per gram (J/g) and kilojoules per kilogram (kJ/kg). Specific latent heat is an escalated property of issue
- The specific latent heat of fusion of water is 334000 J/kg. Solution: The energy for a change of state equals the mass multiplied by the specific latent heat. In this case, m = 0.5 kg. L = 334000 J/kg. Problem 2: Calculate the energy required to convert 0.15kg of ethanol from a liquid to a vapor. The specific latent heat of vaporization of.
- Click hereto get an answer to your question ️ What is the amount of heat energy required to convert ice of mass 20 Kg at - 4^0C to water at 20^0C ? Use the relevant option for calculation.a) Latent heat of fusion of ice 3.34 x 10^5 J / Kg.b) Specific heat capacity of water is 4180 J / Kg / Kc) Specific heat of ice 2093 J / Kg /
- Given specific heat capacity of ice is 2100 J kg-K, specific heat capacity of water is 4186 J kg K, latent heat of fusion of ice is 3.35 x 1053 kg-- and latent heat of steam is 2.256 x10®) kg-. 10

- What mass (in grams) of steam at 100°C must be mixed with 160 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 15.0°C? The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg, and the latent heat of vaporization is 2256 kJ/kg. please help me Ive been trying for so long pleas
- L is specific latent heat for a particular matter (kJ kg-1); Lv for vaporization and Lf for fusion Note: The latent heat of water at 0 degree Celsius for fusion is nearest to 334 joules per gram or 79.7 calories per gram
- The specific latent heat of fusion of water is 334000 J/kg. Solution: The energy for a change of state equals the mass multiplied by the specific latent heat
- The latent heat of fusion for water, L sub f, is 33.5 * 10^4 J/kg. Since the heat lost by the soda is the only heat available to melt the ice, the heat transferred Q is the same in each of our.

Latent heats of fusion and vaporisation The specific latent heat of fusion is the heat required to change 1 kg of a substance from the solid state to the liquid state (or vice versa) at constant temperature 3. Calculate the latent heat of fusion L for each of the three trials you performed. Assume that the energy lost by the warm water is gained by the ice, but notice that the energy is used ﬁrst to melt the ice, then to raise the temperature of the melted ice to match the water (the ﬁnal temperature of water + melt ice is the same) Specific heat of fusion is the heat of fusion to be added per unit mass of a solid to be melted! In the case of ice, the latent heat to be added for melting is 334 kJ per kilogram, which is about the same as the amount of heat that would be needed to bring the water to a boil from room temperature The specific latent heat of fusion of a substance lf is the quantity of heat required to change a unit mass of a solid to a liquid without a change in the temperature. This occurs at the melting..

Heat of Fusion-the amount of heat required to convert unit mass of a solid into the liquidwithout a change in temperature. (or released for freezing) For water at its normal freezing point of 0 ºC, the specific heat of Fusion is 334 J g-1 How much heat must be removed from 456 g of water at 25.0°C to change it into ice at -10.0°C? The specific heat of ice is 2090 J/kg • K, the latent heat of fusion of water is 33.5 ¥ 104 J/kg, and the specific heat of water is 4186 J/kg • K

- Find the heat needed to convert 10 g of water into vapour (Latent heat of vapourisation of water = 2. 2 4 × 1 0 6 J k g − 1) View Answer Define Latent heat of Fusion
- Latent heat of fusion of ice = 3.34 X 10 5 joule / kg specific heat of water = 4.180 X 10 3 j/kg- C. Solution: In this case heat supplied is used into two parts. To change the state of ice into the water at 0 C; To increase the temperature of water from 0 C to 25 C. Heat required to change ice into water at 0 C. Q 1 = mL. Here m = 1.5kg, Latent.
- g up. 'Qsolid = 'Qphase change + 'Qtemperature change = mLf + mc'T (equ.1) Where: m: is the mass of the solid (in our case ice) in Kg L f: is the latent heat of fusion of water in J/Kg c: is the specific heat of water in J.
- Assume the latent heat of fusion of the water is 80 kcal/kg and that the specific heat of water is 1 kcal/ (kg o C)

- Latent heat (also known as latent energy or heat of transformation) is energy released or absorbed, by a body or a thermodynamic system, during a constant-temperature process — usually a first-order phase transition.. Latent heat can be understood as energy in hidden form which is supplied or extracted to change the state of a substance without changing its temperature
- [Specific heat capacity of water = 4200 J kg-1 °C-1] [Specific latent heat of fusion of ice = 336 × 10 3 J kg-1] Solution: Let mass of water used = m gm. Question 7: 250 g of water at 30 °C is present in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5 °C
- Latent heat of fusion of ice=3.36 × 105 J/kg and latent heat of vaporization of water= 2.26×106 J/kg. Solution 7 Heat released by steam to get converted into water=msteamL= (1) (2.26×106

Water has a high latent heat of fusion, so turning water into ice requires the removal of more energy than freezing liquid oxygen into solid oxygen, per unit gram. Latent heat causes hurricanes to intensify. Air heats as it crosses warm water and picks up water vapor. As the vapor condenses to form clouds, latent heat is released into the. Similarly, while ice melts, it remains at 0 °C (32 °F), and the liquid water that is formed with the latent heat of fusion is also at 0 °C. The heat of fusion for water at 0 °C is approximately 334 joules (79.7 calories) per gram, and the heat of vaporization at 100 °C is about 2,230 joules (533 calories) per gram (Specific latent heat of fusion of water = 3,34,000J/kg, Specific heat capacity of water = 4200Jkg-1 K-1). heat; class-9; Share It On Facebook Twitter Email. 1 Answer +1 vote . answered Oct 26, 2020 by Naaji (56.7k points) selected Oct 26, 2020 by Jaini . Best answer. Total heat = Heat required to convert 2Kg of ice into water at 0°C + Heat. = 2260 kJ/kg. Latent Heat of Fusion of Water: For water at its normal freezing temperature or melting point (0°C), the latent heat of fusion is. L = 80 cal/ g = 60 kJ/mol = 336 kJ/kg. It is more painful to get burnt by steam rather than by boiling water at 100°C. Steam converted to water at 100°C, then it gives out 536 cal of heat, so, it is.

- How much heat must be removed from 456 g of water at 25.0°C to change it into ice at -10.0°C? The specific heat of ice is 2090 J/kg ∙ K, the latent heat of fusion of water is 33.5 × 104 J/kg, and the specific heat of water is 4186 J/kg ∙ K. Physics. A 95.1-g ice cube is initially at 0.0°C
- utes. Assume that the specific latent heat of fusion of the solid is 95 000 J/kg and that heat exchange with the surroundings may be neglected.
- Favorite Answer The latent heat of fusion for water is 3.35 × 10^5 J/kg, while the latent heat of vaporization is 2.26 × 10^6 J/kg. What mass m of water at 0 °C must be frozen in order to release..
- There are two types of latent heat works for water as follows- 1st- latent heat of vaporization # Heat required to change the phase of water from liquid to gas and gas to liquid without changing its temperature. The latent heat of vaporization of.
- (The first is used to extract the heat of combustion directly.) This cools the exhaust gases, but more importantly it condenses the water vapor to extract its precious latent heat. This allows furnaces that would normally extract 80 to 85% of the heat of combustion to reach efficiencies of 90 to 95%
- Latent Heat of Fusion and Vaporisation. The energy required to change the phase of a substance is known as a latent heat. The word latent means hidden. When the phase change is from solid to liquid we must use the latent heat of fusion, and when the phase change is from liquid to a gas, we must use the latent heat of vaporisation
- Latent Heat of Fusion. In case of solid to liquid phase change, the change in enthalpy required to change its state is known as the enthalpy of fusion, (symbol ∆H fus; unit: J) also known as the (latent) heat of fusion. Latent heat is the amount of heat added to or removed from a substance to produce a change in phase

** The latent heat of fusion or melting of ice of mass 1kg is 336000J **. This means that 1kg ice requires 336000J of heat to change to water without any change in the temperature. This in turn happens because the heat energy absorbed brings about chan.. That is 334kJ/kg. What is the latent heat of vaporization of water? That is 2264.705kJ/kg. Let us calculate latent heat by a simple expression: Determine the latent heat of a 10kg substance if the amount of heat for a phase change is 200k.cal. By using the formula, Given that Mass (M) = 10 kg, Amount of heat (Q) = 200k.cal. Latent heat = Q/M. Latent heat is measured in units of J/kg. Both L f and L v depend on the substance, particularly on the strength of its molecular forces as noted earlier. L f and L v are collectively called latent heat coefficients.They are latent, or hidden, because in phase changes, energy enters or leaves a system without causing a temperature change in the system; so, in effect, the energy is hidden The latent heat for the water/steam transition is 2256 kJ/kg, to convert 1 kg of water into steam.. In this equation, ΔQ is the amount of heat ﬂow, as before, m is the mass of the object, and L is an intrinsic constant of the material. If the material is melting (i.e. changing from a solid to a liquid), then L is known as the Latent Heat of Fusion and is written as L f . If the object is being vaporized (i.e. changing from a liquid t

Q. How many Joules of energy are required to change 0.010 kg of ice at -2 ο C to water at 20 ο C? The specific heat capacity of water is 4181 J / kg °C. The specific heat capacity of ice is 2090 J / kg °C. The specific latent heat of fusion of ice is 334000 J/kg Q = heat supply = 863 KJ = 863000 J. conversion used : (1 KJ = 1000 J) m = mass of the substance = 4.6 Kg = 4600 g. conversion used : (1 Kg = 1000 g) = specific latent heat of fusion = ? Now put all the given values in the above formula, we get the specific latent heat of fusion of ice. Therefore, the specific latent heat of fusion of ice is.

( Specific heat capacity of ice = 2.1 x 103J kg-1oC-1, Specific heat capacity of ice = 4.2 x 103J kg-1oC-1, Latent heat of fusion of ice = 3.34 x 105 J kg-1 and Latent heat of vaporisation of water = 2.26 x 106 J kg-1) Solution To determine the latent heat of fusion of ice The electrical power of the heater is recorded = P The mass of each the. The latent heat of fusion or melting of solid is the quantity of heat in joules required to convert 1 kg of solid to liquid,without any change in temperature. The latent heat of fusion of ice is 3.34 ⨰ 105 /kg. For Ex:latent heat of vaporisation of water is 22.5 ⨰105 J/Kg Water in a pan reaches 1 0 0 ∘ C, but the pan is still left on the heat, so eventually all of the water turns to water vapor. Calculate the energy needed to evaporate the 1.2 kg of water contained by the pan. Use a value of 2,258 kJ/kg for the specific latent heat of vaporization of water. Give your answer to the nearest kilojoule

An ice machine removes heat at the rate of 4.5 kW. How long does it takes to freeze 4 kg of water at the temperature 14°C? [Specific latent heat of fusion of ice = 3.36 × 10 5 J kg −1, specific heat capacity of water = 4.2 × 10 3 J kg −1 °C −1 heat. Latent Heat Energy must be removed from substance Example How much heat is released from 50g of water as it (a) Changes from liquid to ice at 0oC. (b) Changes from steam to liquid water at 100oC. Q = mL f Q = mL v = 0.05kg*335 kJ/kg = 16,750J = .05kg*2260kJ/kg = 113,000J Steam to water transition liberates more heat than water-ice.

If you want to find the latent heat of fusion, you take a cup with 0.755 kg of water at 30.00 C and drop 0.0850 kg of ice initially at 0.000 C and it all reaches an equilibrium temperature of 19.00 C, what is Lf? (the specific heat of water is 4186 J/(kg C) 2. Find the energy needed to change 500 g of ice at 0 o C to water at 0 o C. Specific latent heat of fusion of water = 334 000 J/kg. Energy required = 0.5 x 334 000 = 167 000 J. Much more heat is needed to turn 1 kg of water into steam than to melt 1 kg of ice The specific heats of ice and water are cI = 2.10×103 J/(kg⋅°C) and cW = 4.19×103 J/(kg⋅°C), respectively, and the latent heat of fusion for water is Lf = 3.34×105 J/kg. Enter an expression for the mass of ice you added, in terms of the defined quantities

Temperature of the each compartment remain homogeneous during whole **heat** transfer process. Given specific **heat** **of** ice = 2100 J/kg-K Given specific **heat** **of** **water** = 4200 J/kg-K **Latent** **heat** **of** **fusion** **of** ice = 3.3 × 105 J/kgQ.The value of rate P isa)42.0 Wb)36.0 Wc)21.0 Wd)18.0 WCorrect answer is option 'A'

19c Latent Heat of Fusion - 3 - 9. Use Equation 2 and the data in the Data Section to solve for the latent heat of fusion Lf. Heat lost = Heat gained Heat lost by warm water = Heat needed to melt ice + Heat needed to warm the water which was once ice M wCw (T w - T f) = M ice Cw (T f - 0) + M ice Lf Eq. Object of the Experiment Energy is required to change water from a solid to a liquid, i.e. to melt ice. In this experiment you will try to measure the latent heat of fusion of ice (LHice), the energy needed (per gram) to melt ice. The needed energy will come from a cup of warm water L = specific latent heat of fusion of substance that is measured in Joule per kilogram or J/kg. Conclusion. A matter transforms from one state to another because of a hidden heat, which is latent. If energy is added to rise by 1 kg, it is the specific latent heat of fusion. Both are measured in J/kg The latent heat of fusion of ice = 3.34 x 10 5 J/kg. The latent heat of fusion of water = 80 cal/gram. Ice at 0 degrees Celsius is more effective in cooling a substance than water at 0 degrees Celsius. If a liquid freezes to form a solid, an equal amount of heat is given out

What is the latent heat of vaporization of water? That is 2264.705kJ/kg. Let us calculate latent heat by a simple expression: Determine the latent heat of a 10kg substance if the amount of heat for a phase change is 200k.cal. By using the formula Water has a very high latent heat for evaporation and fusion, due to the forces between molecules associated with hydrogen bonds (2.5 x 106J/kg). Ice is highly ordered, less dense than liquid water (very unusual), and has significant latent heat of fusion (0.34 x 106J/kg) (d) A body of mass 2 kg at 100 oC is dropped gently into a mixture of ice and water at 0oC. Calculate the mass of ice that melted at 0oC. Specific latent heat of fusion of ice = 3.3 x 105 J kg-1 Specific heat capacity of the body = 2.2 x 103 J kg-1 K- The latent heat of fusion of ice is 334 kJ/kg, the specific heat of ice is 2050 J/(kg·°C), and the specific heat of water is 4180 J/(kg·°C). Calculate the final temperature of the system if 383.9 kJ of energy is added by heat. Answer: 57 °C. 6) A system of 480 g of water is initially at 19 °C. The latent heat vaporization of water is 2260.

The specific heat of ice is 2090 J/(kg K) and the latent heat of fusion of water is 33.5 × 104 J/kg. A)546 g B)302 g C)345 g D)405 g E)634 g 24)An 920-g piece of iron at 100°C is dropped into a calorimeter of negligible heat capacity containing 50 g of ice a 7 Explain why the value for the specific latent heat of vaporisation of water is greater than the value for the specific latent heat of fusion of water. 8 Explain why: a specific heat capacity can be stated in J/kg K or J/kg °C with the same value b specific latent heat has units of J/kg and not J/kg °C Specific latent heat is measured in J/kg, if energy is measured in J and mass in kg.For example, specific latent heat of ice is 334000J/kg means 334000 J of energy is needed to convert 1kg of water into ice or vice versa

Just as latent heat is taken in when water changes to vapour at the same temperature, so the same thing occurs when ice melts to form water. But in this case the latent heat is not so great. It requires only 336 000 J to convert 1 kg of ice at 0 °C to water at the same temperature Latent heat is an intensive property measured in units of J/kg. Both Lf and Lv depend on the substance, particularly on the strength of its molecular forces as noted earlier. Lf and Lv are collectively called latent heat coefficients Using the equation for the heat required for melting, and the value of the latent heat of fusion of water from the previous table, we can solve for part (a). Solution to (a) The energy to melt 1.000 kg of ice i

Determine the energy required to melt all of the ice from -20 °C to water at a temperature of 0 °C. Specific latent heat of fusion of ice = 330 kJ kg Specific heat capacity of ice = 2.1 kJ kg k Density of ice = 920 kg m Markscheme mass = «volume x density» (0.75) x 920 «= 388 kg Latent heat is an intensive property measured in units of J/kg. Both L f and L v depend on the substance, particularly on the strength of its molecular forces as noted earlier. L f and L v are collectively called latent heat coefficients. They are latent, or hidden, because in phase changes, energy enters or leaves a system without causing a temperature change in the system; so, in effect, the. Calculate the Specific Latent Heat of Fusion of Ice. Specific Heat Capacity of Water is `4200 J Kg^-1 K^-1` - Physics 10g of ice at 0℃ absorbs 5460 J of heat energy to melt and change to water at 50℃. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is `4200 J kg^-1 K^-1

Water has a very high latent heat for evaporation and fusion, due to the forces between molecules associated with hydrogen bonds (2.5 x 106 J/kg). Ice is highly ordered, less dense than liquid water (very unusual), and has significant latent heat of fusion (0.34 x 106 J/kg) where the latent heat of fusion, L f size 12{L rSub { size 8{f} } } {}, and latent heat of vaporization, L v size 12{L rSub { size 8{v} } } {}, are material constants that are determined experimentally. See ().Latent heat is measured in units of J/kg

Latent heat of vaporization - water at 3 MPa (pressure inside a steam generator) hlg = 1795 kJ/kg Latent heat of vaporization - water at 16 MPa (pressure inside a pressurizer) hlg = 931 kJ/k the explanatory text, the values used by the simulator for the specific heat of water, the specific heat of ice, and the latent heat of fusion of ice. Compare the value of melted ice to your calculated value. 3) If the initial mass of the water is , using equation (6), calculate the amount of ice expected to melt Latent heat of fusion of ice=3.36 × 10 5 J/kg and latent heat of vaporization of water= 2.26 × 10 6 J/kg. Solution 7 Heat released by steam to get converted into water=m steam L 푣 =(1)(2.26 × 10 6 ii) The specific latent heat of fusion of ice is 336000 J kg-1. iii) The specific heat capacity of copper is 0.4 J g -1o C -1 i) It means 50 J of heat is required to raise its temperature by 1 o [Specific latent heat of fusion of water = 334 000 J/kg; specific heat capacity of water = 4200 J/ (kg °C).] Energy needed to melt 2kg of ice, Q1 = mL = (2) (334000) = 668000J Energy needed to change the temperature from 0°C to 20°C

some juice at +10 oC. Suppose the specific heat of the juice is 4000 J/(kg oC), the specific heat of ice is 2000 J/(kg oC) and the latent heats of fusion of water and the juice are the same, equal to 3 x 105 J/kg. Answer the following questions based on the information provided. You may assume that the cup Mary uses in making the drink. The water is heated until it boils at #100^0C#.Given that the specific heat capacity of water =4200 J/Kg/K, heat capacity of the kettle = 450 J/K, specific latent heat of vaporization of water =2.3mJ/Kg. 5 Substance: Melting point K: Melting point °C: Heat of fusion (10 3 J/kg): Helium: 3.5-269.65: 5.23: Hydrogen: 13.84-259.31: 58.6: Nitrogen: 63.18-209.97: 25.5. Deduce a value for the specific latent heat of fusion of ice. 336000J kg-1; A mixture of ice and water at 0 °C has a total mass of 1 kg, and contains 0.5 kg of ice. An immersion heater melts all the ice in 168 s, and then needs a further 84 s to heat the mixture to 20 ° C. Find the specific latent heat of fusion of ice. 336000 Jkg-1; A.

Specific **latent** **heat**• **Latent** **heat** **of** **fusion** refers to the change from asolid to a liquid (melting)• **Latent** **heat** **of** vaporisation refers to the changefrom a liquid to a gas (boiling).• Example 1 - the specific **latent** **heat** **of** **fusion** **of** iceis 330 000 J kg-1 (II) Determine the latent heat of fusion of mercury using the following calorimeter data: 1.00 kg of solid Hg at its melting point of -39.0^\circC is placed in Our Discord hit 10K members! Meet students and ask top educators your questions The measurement techniques presently used for latent heat of fusion and melting m2 Cp Specific heat, J/kg o K dt Time interval, s Gr Grashof number h Heat transfer coefficient, W/m 2 o K hcw. The specific heat of water = 4200 J kg -1 K -1 and the latent heat of ice = 3.4 × 10 5 J kg -1. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) (1) 69.3 (2) 63.8 (3) 64.6 (4) 61.

Physics P Worksheet 12.2 Latent Heat, Specific Heat, and Work Worksheet 12.2 How much dry ice (heat of fusion = 184 kJ/kg) needs to be added to 400 g of water at 20˚C to cool the water to 0˚C. Assume that once the CO 2 sublimes it leaves the water and absorbs no additional heat. 11. How far does a 1 g hailstone at 0˚C need to fall in. Latent heat is energy released or absorbed, by a body during a constant-temperature process, for example, a phase change of water from liquid to gas. This is written as: Sensible heat = (mass of the body) * (specific latent heat) The equation is w.. The specific heat of mercury is 138 J/kg° C. Determine the latent heat of fusion of mercury using the following calorimeter data: 1.00 kg of solid Hg at its melting point of -39.0° C is placed in a .620-kg aluminum calorimeter with 0.400 kg of water at 12.80° C; the resulting equilibrium temperature is 5.06° C If the specific latent heat of ice is 3.3 x 10 5 J/kg and that of steam is 2.3 x 10 6 J/kg and the Sp. heat capacity of water is 4200 J/kg, draw rough graphs for the following to illustrate the changes which take place as the solid ice is converted into water and finally into steam, the pressure remaining constant all the time Simple examples of latent heat: When we boil the water and the temperature remains at 100°C until the last drop evaporates, due to the latent heat of vaporization, the added heat in the water is absorbed and carried away by releasing vapor molecules. Latent heat of fusion of water: That is 334kJ/kg -Latent heat of fusion = 6,003 J/mol = 333.5 kJ/kg •Liquid nitrogen -At atmospheric pressure, boiling point is ‐195.8 °C -Latent heat of vaporization = 5,580 J/mol = 199 kJ/kg •Dry ice (solid CO2) -At atmospheric pressure, sublimation point is ‐78.5 °