Question

A copper pot with a mass of 0.485 kg contains 0.170 kg of water, and both are at a temperature of 18.5 ∘C. A 0.225 kg block of iron at 88.0 ∘C is dropped into the pot.

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

Answer 1

Here,

let the final temperature of the system is T

heat lost by iron = heat gain by the copper + water

0.225 * 450 *(88 - T ) = 0.485 * 350 * (T - 18.5) + 0.170 * 4186 * (T - 18.5)

solving for T

T = 25.7^o C

the final temperature of the system is 25.7^o C