Question

A copper pot with a mass of 0.475 kg contains 0.185 kg of water, and both are at a temperature of 19.0 ∘C . A 0.240 kg block of iron at 86.5 ∘C is dropped into the pot.Find the final temperature of the system, assuming no heat loss to the surroundings.

Answer 1

Answer:

The final temperature is 25.85°C.

Explanation:

Given that,

Mass of copper pot = 0.475 kg

Mass of water = 0.185 kg

Temperature = 19.0°C

Mass of iron block = 0.240 kg

Temperature = 86.5°C

We need to calculate the final temperature

Using formula of heat

Heat lost by iron block=Heat gained by copper pot and water

$-M_{i}C_{i}\Delta T=M_{w}C_{w}\Delta T+M_{c}C_{c}\Delta T$

Put the value into the formula

$-0.240\times450\times(T-86.5)=0.185\times4180\times(T-19.0)+0.475\times384\times(T-19.0)$

$-108\timesT+108\times86.5=773.3\times T-773.3\times19.0+182.4\times T-182.4\times19.0$

$9342-108T=773.3T-14692.7+182.4T-3465.6$

$-108T-773.3T-182.4T=-9342-14692.7-3465.6$

$1063.7T=27500.3$

$T=\dfrac{27500.3}{1063.7}$

$T=25.85^{\circ}C$

Hence, The final temperature is 25.85°C.