Question

$5g$ of steam at ${100}^{\circ }C$ is mixed with $10g$ of ice at $0^{\circ }C$. At equilibrium, the mixture contains $\frac{p}{q}$ gram of steam. Find $(p+q)$.

(Given: $s_{\text{water}}=1cal/g^{\circ }C,L_f=80cal/g,L_v=540cal/g$)

Answer 1

We know that: $Q=mL,Q=ms\Delta \theta$

Given: $s_{\text{water}}=1cal/g^{\circ }C,L_f=80cal/g,L_v=540cal/g$

Available heat : $5g\text{steam}\left({100}^{\circ }C\right)$

$Q=mL=5\times 540=2700cal$

Required heat : $10gice\left(0^{\circ }C\right)$

$Q=mL=10\times 80=800cal$

$10g\text{water}\left(0^{\circ }C\right)$

$Q=ms(100-0)$

$=10\times 1\times (100-0)$

$=1000cal$

So available heat is more than required heat, therefore final temperature will be ${100}^{\circ }C$.

Mass of steam condensed

$m=\frac{Q}{L}=\frac{800+1000}{540}=\frac{10}{3}g$

Total mass of steam

$=5-\frac{10}{3}=\frac{5}{3}g$

Comparing with $\frac{p}{q}$ we get,

$p=5,q=3$

$p+q=5+3=8$


Final answer: $8$.