Question

Find the amount of heat required to convert $10$g of ice at $-{10}^o$C to steam at ${100}^o$C.

Answer 1

10g of ice at -10℃ $⟶$ steam at 100℃

$\downarrow Q_1$

10g of ice 0℃

$\downarrow Q_2$ $\uparrow Q_4$

10g water 0℃ $⟶Q_3$ 10g water 100℃

First we need to bring the ice at -10℃ to 0℃.

$△$T = 10

m = 2g

S = 1cal/g/℃

$Q_{11}$ = ms$△$T

$\Rightarrow$ = 10 cal.

Then, we convert this ice into water Lfusion = 80cal/g.

$Q_{12}$ = mLfusion

$\Rightarrow$ = 2$\times$80

$\Rightarrow$ = 160cal

Then we can increase the temp. of this water to 100℃

$△$T = 100-0

$\Rightarrow$ = 100

$Q_{13}$ = ms$△$T

$\Rightarrow$ = 2$\times$100$\times$1

$\Rightarrow$ = 200cal.

Now, we convert this water into steam.

Lvap = 540cal.

$Q_{14}$ = mLvap

$\Rightarrow$ = 2$\times$540

$\Rightarrow$ = 1080cal

Heat required = $Q_{11}+Q_{12}+Q_{13}+Q_{14}$

$\Rightarrow$ = 10 +160+200+1080

$\Rightarrow$ = 1450cal.