Question

Ice at 0°C and 1 atm has a density of 916.23 kg/m3, while water under 

these conditions has a density of 999.84 kg/m3. How much work is 

done against the atmosphere when 10 kg of ice melts into water?

Answer 1

Answer

Volume of Ice

$= \frac{10Kg}{916.23 Kg/m^3 }\\= 1.0914 \times10^{-2 }m^3$

When the Ice melts, Its density changes to 999.84 Kg/m^3 .

Volume of water formed

$= \frac{10Kg}{999.841}\\ = 1.00016 \times10^{-2} m^3$

Also atmospheric pressure

$= 1.01\times10^5 Pa$

which is constant throughout

Now work done at constant pressure is given by $W=P \times (V_2 - V_1).$

So in this case work done

$W= 1.01 \times 10^5 \times (1.00016 - 1.0914) \times 10^{-2}\\ = -92.15 Joules$