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* 0.12 * (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* 1 * (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* 0.12 * (40-T)=100(T-20)`

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

`\Rightarrow 0.12* 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`.