Solution:
According to the question, 10 g of ice at -10oC is mixed with 10 g of water at 10oC.
And, heat capacity of ice = 2.1 J g-1 K-1,
specific latent heat of ice = 336 J g-1, and
specific heat capacity of water = 4.2 J g-1 K-1
Consider that whole of the ice melts and the final temperature of the mixture is T0 C. Then by using the expression for het energy => Q = m × c × (change in temeprature), we can write –
Heat energy gained by 10 g of ice at – 100 C in order to raise its temperature to 00 C
= 10 × 10 × 2.1 = 210 J
Similarly, the heat energy gained by 10 g of ice at 00 C in order to convert into water at 00 C
= 10 × 336 = 3360 J
Again, the heat energy gained by 10 g of water (obtained from ice) at 00 C to raise its temperature to T0 C
= 10 × 4.2 × (T – 0) = 42 T
Now, the heat energy released by 10 g of water at 100 C in order to lower its temperature to T0 C becomes
= 10 × 4.2 × (10 – T) = 420 – 42T
Since, there is no energy loss, heat energy gained is equal to the heat energy lost
Therefore, equating total heat energy gained and lost in the mentioned processes, we get
210 + 3360 + 42 T = 420 – 42T
or, T = -37.50 C
Water cannot exist at this temperature, hence this can’t be true. As a result, the ice does not melt completely. Allow m gram of ice to melt. The mixture reaches a final temperature of 0 degrees C. As a result, the quantity of heat energy obtained by 10 g of ice at -100 C to raise its temperature to 0 degrees C is
= 10 × 10 × 2.1 = 210 J
Heat energy gained by m gm of ice at 0 C to convert to water at 0 C = m ×336 = 336 m J
Then, heat energy emitted by 10 g of water at 100 C to drop its temperature to 0 degrees C =
10 × 4.2 × (10 – 0) = 420
We know that the heat energy gained equals heat energy lost. So,
210 + 336 m = 420
Or, m = 210 / 336
Thus, m = 0.625 gm