Question

12. 500 g of ice at 0°C is mixed with 1 g steam at 100°C. The final temperature of the mixture is (1) 0°C (2) 50°C (3) 40°C (4) 100°C ?​

Answer 1

Answer:

0°C

Explanation:

Mass of Ice = 500 gm

Mass of Steam = 1 gm

Temperature of Ice = 0°C

Temperature of Steam =100°C

Let us assume the common interface to be water at 0°C

Therefore heat energy required to convert ice at 0°C into water at 0°C = m*L

(Where 'm' is mass of substance and 'L' is latent heat)

Q = 500 * 80 => Q = - 40000 cal

(Negative sign denotes heat energy is required)

Now, let us find heat released when Steam at 100°C convert to water at 0°C

Q = msT + mL

(Where 's' is Specific Heat Of Water and 'T' is change in temperature)

Q = 1*1*100 + 1*540 => Q = 100 + 540

Q = + 640 cal

(Positive sign denotes heat energy is released)

Net Q = - 40000 + 640 => Q = - 39360 cal

M = Q/L

(Where 'M' is mass of substance that has changed its phase)

M = 39360/80 => M = 492 gm

This means that some all of the heat released by steam melted 8gm of Ice and now a mixture of of Ice and water exists.

This is possible only at melting point of Ice, i.e. at 0°C

Hope this answer helps you...