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Question 1
Calculate the energy released when
(a) 10 g water at 100 °C and
(b) 10 g of steam at 100 °C
are 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. |
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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 is
E = ml = 0.01 × 2.2 × 106 = 22 000 J
The heat given out when this condensed water drops in temperature from 100 °C to 33 °C is
E = mcq = 0.01 × 4200 × (100 – 33) = 2814 J
So, the total heat given out is = 25 kJ |
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Question 2
When 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 |
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Definition
Solution
Potential energy = mgh = 2.5 × 10–3 × 9.81 × 2 × 103
= 49.05 J
The falling hailstone loses potential energy, and this is used to partly melt the hailstone.
ml= 49.05
m × 3.36 × 105 = 49.05
m = 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 |
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Question 3
0.30 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 . |
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Definition
Solution
Let q be the final temperature.
Heat lost by water = heat gained in melting the ice + heat gained in warming the ice water
mwcwDqw = micelice + micecwDqmelted ice
mwcw(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 |
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Question 1
Given 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. |
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Definition
Solution
Heat lost by copper = heat gained by water
mcuccuDqcu = mwcwDqw
mcuccu(100 - 24) = 0.200 × 11ccu(24 – 20)
76mcu = 8.8
mcu = 0.116 kg |
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| 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 |
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Definition
Solution
Let 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 |
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Question 3
1 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. |
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Definition
Solution
Let 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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