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

The final temperature is 25.85°C.

Step-by-step 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*450*(T-86.5)=0.185*4180*(T-19.0)+0.475*384*(T-19.0)`

`-108*T+108*86.5=773.3* T-773.3*19.0+182.4* T-182.4*19.0`

`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=(27500.3)/(1063.7)`

`T=25.85^(\circ)C`

Hence, The final temperature is 25.85°C.