10 g of ice at −10°C is mixed with 10 g of water at 0°C. The amount of heat required to raise the temperature of mixture to 10°C is
- A.400 cal
- B.550 cal
- C.1050 cal
- D.1200 cal
Correct answer
Explanation
The correct answer is C, 1050 cal. The heat needed is 50 cal to warm the ice to 0°C, 800 cal to melt it, and 200 cal to warm all 20 g of water to 10°C.
Take the specific heat of ice as 0.5 cal/g°C, the latent heat of fusion of ice as 80 cal/g and the specific heat of water as 1 cal/g°C. Warming 10 g of ice from −10°C to 0°C needs 10 × 0.5 × 10 = 50 cal. Melting it at 0°C needs 10 × 80 = 800 cal, and this is the largest part, since the temperature does not rise while ice melts. The mixture is then 20 g of water at 0°C, and raising it to 10°C needs 20 × 1 × 10 = 200 cal. The total is 50 + 800 + 200 = 1050 cal.
A, 400 cal, is only half of the 800 cal needed just to melt the ice.
B, 550 cal, is also too small, since melting the ice alone takes 800 cal.
D, 1200 cal, is more than the three steps add up to; even taking the specific heat of ice as 1 cal/g°C gives only 1100 cal.
Exam tip: specific heat of ice 0.5, of water 1 cal/g°C; latent heat of fusion 80 cal/g, of vaporisation about 540 cal/g.
Practice Questions
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Show answer
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