Question

A copper pot with a mass of 0.500 kg contains 0.170 kg of water, and both are at a temperature of 20.0 ∘C . A 0.250 kg block of iron at 85.0 ∘C is dropped into the pot.

Find the final temperature of the system, assuming no heat loss to the surroundings.

Answer 1

Concepts and reason

The concept used to solve this problem is the heat equation in thermodynamics.

Initially, use the concept of equilibrium temperature of the system. Then, find the final temperature of the system by equating the heat lost by the iron block with the heat gained by the copper and the water. Finally, rearrange the expression for the temperature and then calculate.

Fundamentals

From the zeroth law of thermodynamics when two objects are places in contact heat is transfer from high temperature to low temperature until they reach the same temperature.

The iron block has higher temperature than copper pot contain water. Hence the temperature will transfer from iron block to copper pot and water.

The expression for the equilibrium temperature of the system is,

Here, is the heat lost by the iron block, is the heat gained by water, and is the heat gained by copper pot.

The expression for the amount of heat required to change the temperature is,

Here, is the amount of heat, is the mass, is the specific heat, and is the change in temperature.

The expression for the amount of heat required to change the temperature is,

Expression for the amount of heat lost by the iron block is,

Here, is the mass of iron, is the specific heat of iron, and is the temperature difference in the iron.

Assume that is the equilibrium temperature.

Substitute for , for , and for .

The heat lost is,

…… (1)

Expression for the amount of heatgained by water is,

Here, is the mass of water, is the specific heat of water, and is the temperature difference in the water.

Assume that is the equilibrium temperature.

Substitute for , for , and for .

The heat gained by the water is,

…… (2)

The expression for the amount of heat gained by copper is,

Here, is the mass of copper, is the specific heat of copper, and is the temperature difference in the copper.

Assume that is the equilibrium temperature.

Substitute for , for , and for .

The heat gained by the copper is,

…… (3)

The expression for the equilibrium temperature of the system is,

Substitute for , for , and for .

Solve the above equation for .

Therefore, the final temperature of the system is .

Ans:

The final temperature of the system is .