Question: 10 kg of ice at –5<sup>0</sup>C is kept in a large container of negligible heat capacity 1 kg of ste...

Question

10 kg of ice at –5⁰C is kept in a large container of negligible heat capacity 1 kg of steam at 110⁰C is passed over the container. The final temp of mixture when thermal equilibrium is attained :

[Specific Heat of ice = 0.5 cal/gm⁰C;

Specific Heat of water = 1 cal/gm⁰C;

Specific Heat of steam = 0.47 cal/gm⁰C ;

Latent Heat of fusion = 80 cal/gm

Latent Heat of vaporization = 540 cal/gm]

Answer 1

Solution

Heat required by ice to melt

̃ $\left\lbrack 10000 \times \frac{1}{2} \times 5 \right\rbrack$ + [10000 × 80]

̃ 25000 + 800000

̃ 825000 Cal.

Heat lost by steam if it finally reaches upto even 0⁰.

̃ [1000 × 0.47 × 10] + [1000 × 540]

[1000 × 1 × 100]

̃ 4700 + 540000 + 100000

̃ 644700 Cal

Comparing both final temperature of mixture will be 0⁰C.