Application of heat capacity 2

Classical Mechanics Level 2

Suppose 61.0 61.0 g hot metal, which is initially at 120.0 120.0 °C, is plunged into 100.0 100.0 g water that is initially at 20.00 20.00 °C. The metal cools down and the water heats up until they reach a common temperature of 26.39 26.39 °C. Calculate the specific heat capacity of the metal, using 4.18 4.18 J K − 1 ^{-1} g − 1 ^{-1} as the specific heat capacity of the water.

0.468 0.468 J K − 1 ^{-1} g − 1 ^{-1} 0.936 0.936 J K − 1 ^{-1} g − 1 ^{-1} 0.532 0.532 J K − 1 ^{-1} g − 1 ^{-1} 0.638 0.638 J K − 1 ^{-1} g − 1 ^{-1}

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Krishna Singh by Krishna Singh , 22, India

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1 solution

Anirudh Krishna
Nov 25, 2015

Heat Lost By the Hot Body = Heat lost by the Cold Body ( Provided No Energy is lost to the surroundings )

Hence, here, the metal body is hotter than the water. Therefore, heat lost by the metal body = heat gained by water. Once both the body and water reach thermal equilibrium, that is, there is no temperature difference between the bodies, heat transfer will no longer take place. Here, the equilibrium temperature is given to be 26.39 deg. C.

Now, Heat Lost/Gained by an object is given by, Q = m x s x t : where m is the mass of the body under observation, s is the specific heat of the body and t is the change in temperature.

Hence, in this problem, m1 x s1 x t1 = m2 x s2 x t2 ( Heat lost by hot body = heat gained by colder body) 100 x 4.18 x (26.39-20.00) = -[61.0 x s2 x (26.39-120.00)] ( - ve signs indicates heat is lost )

Solving which gives, s2 = specific heat capacity of metal to be 0.468.

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