# Shared Flashcard Set

## Details

Latent heat problems
Latent heat problems
7
Physics
10/22/2008

Term
 Question 1Calculate the energy released when(a) 10 g water at 100 °C and(b) 10 g of steam at 100 °Care each spilt on the hand.Take the specific heat capacity of water to be 4200 J kg –1 K –1 and the specific latent heat of vaporisation of water to be 2.2 MJ kg–1 . Assume that the temperature of the skin is 33 °C.
Definition
 Solution(a) E = mcD0 = 0.01 × 4200 × (100 – 33) = 2814 J = 2.8 kJ(b) The latent heat given out in changing from steam at 100 °C to water at the same temperature isE = ml = 0.01 × 2.2 × 106 = 22 000 JThe heat given out when this condensed water drops in temperature from 100 °C to 33 °C isE = mcq = 0.01 × 4200 × (100 – 33) = 2814 JSo, the total heat given out is = 25 kJ
Term
 Question 2When a falling hailstone is at a height of 2.00 km its mass is 2.50 g. What is its potential energy? Assuming that all of this potential energy is converted to latent heat during the fall, calculate the mass of the hailstone on reaching the ground. Take the specific latent heat of fusion of ice to be 3.36 × 105 J kg–1 and the acceleration due to gravity to be 9.81 m s–2
Definition
 SolutionPotential energy = mgh = 2.5 × 10–3 × 9.81 × 2 × 103= 49.05 JThe falling hailstone loses potential energy, and this is used to partly melt the hailstone.ml= 49.05m × 3.36 × 105 = 49.05m = 1.4598 × 10–4 kg (mass of hailstone that melted)Total mass of hailstone = 2.50 g=> Remaining mass that reaches the ground = 2.50 – 0.1458 g= 2.354 g
Term
 Question 30.30 kg of ice at 0 °C is added to 1.0 kg of water at 45 °C. What is the final temperature, assuming no heat exchange with the surroundings? Take the specific heat capacity of water to be 4200 J kg -1 K -1 and the specific latent heat of fusion of ice to be 3.4 × 10 5 J kg -1 .
Definition
 SolutionLet q be the final temperature.Heat lost by water = heat gained in melting the ice + heat gained in warming the ice watermwcwDqw = micelice + micecwDqmelted icemwcw(45 – q) = micelice + micecwD0 melted ice 1 × 4200 × (45 – q) = (0.3 × 3.4 × 105 ) + (0.3 × 4200 × q) 4200 (45 – q) = 1.02 × 105 + 1260 q 1.89 × 105 – 1.02 × 105 = 1260 q + 4200 q q = 16 oC
Term
 Question 1Given that the specific heat capacity of water is 11 times that of copper, calculate the mass of copper at a temperature of 100 °C required to raise the temperature of 200 g of water from 20.0 °C to 24.0 °C, assuming no energy is lost to the surroundings.
Definition
 SolutionHeat lost by copper = heat gained by water mcuccuDqcu = mwcwDqw mcuccu(100 - 24) = 0.200 × 11ccu(24 – 20) 76mcu = 8.8 mcu = 0.116 kg
Term
 Three litres of water at 100 °C are added to 15 litres of water at 40 °C. Calculate the temperature of the mixture. Take the mass of 1 litre of water to be 1 kg and the specific heat capacity of water to be 4.2 × 103 J kg -1 K -1
Definition
 SolutionLet the temperature of the mixture q Heat lost = heat gained m1c1Dq1 = m2 c2 Dq2 3 × 4.2 × 103 × (100 – q) = 15 × 4.2 × 103 × (q -40)Solving for q gives q = 50°C
Term
 Question 31 kg of water at a temperature of 45 °C is mixed with 1.5 kg of alcohol at 20 °C. Find the final temperature of the mixture.Take the specific heat capacity of water to be 4200 J kg –1 K –1 and the specific heat capacity of alcohol to be 2400 J kg –1 K –1 . Assume no other exchange of heat occurs.
Definition
 SolutionLet the final temperature of the mixture be q. Heat lost by water = heat gained by alcohol mwcwDqw = malcalDqal 1 × 4200 × (45 – q) = 1.5 × 2400 × (q – 20) (mcDq)w = (mcDq)al 1 × 4200 × (45 – q) = 1.5 × 2400 × (q – 20)Solving for q gives q = 33 °C.
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