Question

Calculate the specific heat of a metal (in calories/gram-degree C) from the following data. A container made of the metal has a mass of 3.77 kg and contains 18.1 kg of water. A 1.45 kg piece of the same metal, initially at a temperature of 164 degrees C, is placed in the water. The container and water initially have a temperature of 15 degrees C, and the final temperature of the entire system is 18 degrees C.

Answer 1

Answer:

The specific heat of the metal is 0.274951 calories/gram-degree C.

Step-by-step explanation:

Let, the specific heat of the container is x calories/gram-degree C.

The container and water gains (18 - 15) = 3 degrees C.

Hence, the transfer of heat is $3(18.1\times1000\times1) + 3(3.77x\times1000)$.

The metal, which is dropped in the water, losses (164 - 18)= 144 degrees C.

Hence, the transfer of heat is $144\times 1.45\times1000x$.

As per the given conditions,

$3(18.1\times1000) + 3(3.77x\times1000) = 144\times1.45\times1000x\\54.3 + 11.31x = 208.8x\\197.49x = 54.3\\x = 0.274951$.