Question

In lab you submerge 100 g of iron nails at 40 0C in 100 g of water at 20 0 C. Specific heat of iron is 0.12 cal/g∙0C. Equate the heat gained by water to the heat lost by nails and show that the final temperature of water becomes 22.1 0C.

Answer 1

Let T be the final temperature of water and nails (iron).

Given that the initial temperature of 100 g nails, $T_{1F}=40^{\circ}C$.

Specific heat of the iron, $C_F=0.12 cal/g^{\circ}C$.

So, the heat lost by nails $= mC_F(T_{1F}-T}$

$=100\times 0.12 \times (40-T)\cdots(i)$

The initial temperature of 100 g water, $T_{1w}=20^{\circ}C.$

Specific heat of water, $C_w=1 cal/g^{\circ}C$

So, the heat gained by water $= mC_w(T-T_{1w}}$

$=100\times 1 \times (T-20)\cdots(ii)$

As the exergy is conserved, so

the heat lost by the nails = the heat gained by the water

From equation (i) and (ii), we have

$100\times 0.12 \times (40-T)=100(T-20)$

$\Rightarrow 0.12(40-T)=T-20$

$\Rightarrow 0.12\times 40 -120T=T-20$

$\Rightarrow 0.12T+T=4.8+20$

$\Rightarrow 1.12T=24.8$

$\Rightarrow T=24.8/1.12= 22.1^{\circ}C$

Hence, the final temperature of both, iron nails as well as water, become $22.1^{\circ}C$.