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Heat-solutions

# Heat-solutions - le(txl65 Heat mohanty(81440 This print-out...

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le (txl65) – Heat – mohanty – (81440) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points The specific heat of a metal (similar to cop- per) is 0 . 092 cal / g · C. The latent heat of vaporization of a liquid (similar to liquid ni- trogen) is 48 cal / g. A 3 kg block of the metal at 24 C is dropped into a large vessel of the liquid at 77 K, which is the boiling point of the liquid. How many kilograms of the liquid boil away by the time the metal reaches 77 K? Correct answer: 1 . 265 kg. Explanation: Let : m Cu = 3 kg , c Cu = 0 . 092 cal / g · C , T c = 24 C = 293 K , T o = 273 K , T = 77 K , and L N 2 = 48 cal / g . The change in heat is Δ Q = m Cu c Cu ( T Cu - T ) = m N 2 L N 2 m N 2 = m Cu c Cu [ T Cu - T ] L N 2 = (3 kg) (0 . 092 cal / g · C) 48 cal / g × (297 K - 77 K) = 1 . 265 kg . 002 10.0points What is the final equilibrium temperature when 24 g of milk at 24 C is added to 102 g of coffee at 74 C? Assume the specific heats of milk and coffee are the same as that of water, and neglect the specific heat of the container. Correct answer: 64 . 4762 C. Explanation: Let : m m = 24 g , T m = 24 C , m c = 102 g , and T c = 74 C . The amount of heat given out by coffee should be equal to the amount of heat ab- sorbed by milk, so c m m m ( T f - T m ) = c c m c ( T c - T f ) m m T f - m m T m = m c T c - m c T f ( m m + m c ) T f = m c T c + m m T m T f = m c T c + m m T m m m + m c = (102 g) (74 C) + (24 g) (24 C) 24 g + 102 g = 64 . 4762 C . 003 10.0points Two grams of water are sealed in a rigid con- tainer; then the water is vaporized by heating. 976 cal of heat is needed for vaporization. What is the change in internal energy of the water? Correct answer: 4085 . 54 J. Explanation: Let : Q = 976 cal . The water is sealed in the rigid container, so W = 0 and Δ U = Q = (976 cal) · 4 . 186 J 1 cal = 4085 . 54 J . 004 10.0points A 20 kg piece of metal (specific heat of 0 . 06 cal / g · C) having a temperature of 81 C is added to 0 . 7 kg of water having a tempera- ture of 27 C.

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le (txl65) – Heat – mohanty – (81440) 2 What is the final equilibrium temperature of the system? Assume there is no loss of heat to the surroundings. The specific heat of water is 1 cal / g · C. Correct answer: 61 . 1053 C. Explanation: Let : m 1 = 20 kg , c 1 = 0 . 06 cal / g · C , T 1 = 81 C , m 2 = 0 . 7 kg , c 2 = 1 cal / g · C , and T 2 = 27 C . There is no loss of heat to the surroundings, so m 1 c 1 ( T 1 - T f ) = m 2 c 2 ( T f - T 2 ) m 1 c 1 T 1 - m 1 c 1 T f = m 2 c 2 T f - m 2 c 2 T 2 m 1 c 1 T 1 + m 2 c 2 T 2 = ( m 2 c 2 + m 1 c 1 ) T f T f = m 1 c 1 T 1 + m 2 c 2 T 2 m 1 c 1 + m 2 c 2 Since m 1 c 1 T 1 + m 2 c 2 T 2 = (20 kg) (0 .
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