Question

An aluminium vessel of mass 0.5 kg contains 0.2 kg of water at ${20}^{\circ }C$. A block of iron of mass 0.2 kg at ${100}^{\circ }C$ is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium, iron and water are $910K{g}^{-1}{K}^{-1}470Jk{g}^{-1}{K}^{-1}$ and $4200Jk{g}^{-1}{K}^{-1}$ respectively.

Answer 1

Given Mass of aluminium = 0.5 kg

Mass of water = 0.2 kg

Mass of iron = 0.2 kg

Temperature of aluminium and water

$={20}^{\circ }={293}^{\circ }K$

Temperature of iron $={100}^{\circ }C=373k$

Specific heat of Al = 910 J/kg-K

Heat gain $=0.5\times 910(T-293)+0.2\times 4200\times (T-293)$

$=(T-293)$

$[0.5\times 910+0.2\times 4200]$

Heat lost $=0.2\times 470\times (373-T)$

We know Heat gain = Heat lost

$\Rightarrow (T-293)[0.5\times 910+0.2\times 4200]$

$=0.2\times 470\times (373-T)$

$\Rightarrow (T-293)(455+840)=94(373-T)$

$\Rightarrow (T-293)\frac{1295}{94}=(373-T)$ $[\frac{1295}{94}\approx 14]$

$(T-293)\times 14=373-T$

$\Rightarrow 14T-293\times 14=373-T$

$\Rightarrow 15T=373+4102=4475$

$\Rightarrow T=\frac{4475}{15}=298k$

$\therefore T=(298-273{)}^{\circ }C={25}^{\circ }C$

$\therefore$ The final temp $={25}^{\circ }C$