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Home » A piece of ice of mass 40 g is added to 200 g of water at 50oC. Calculate the final temperature of the water when all the ice has melted. Specific heat capacity of water = 4200 J kg-1 K-1 and specific latent heat of fusion of ice =336 x 103 J kg-1.
A piece of ice of mass 40 g is added to 200 g of water at 50oC. Calculate the final temperature of the water when all the ice has melted. Specific heat capacity of water = 4200 J kg-1 K-1 and specific latent heat of fusion of ice =336 x 103 J kg-1.
Calorimetry | Class 10 | ICSE | Physics | Selina
A piece of ice of mass 40 g is added to 200 g of water at 50oC. Calculate the final temperature of the water when all the ice has melted. Specific heat capacity of water = 4200 J kg-1 K-1 and specific latent heat of fusion of ice =336 x 103 J kg-1.

Solution:

According to the question, ice of mass 40 g is added to 200 g of water at 50oC. WE have to calculate the final temperature when all the ice has melted. Let this final temperature of the water be T0 C. First, we will calculate the heat energy involved in the process using the expression given below –

Q = m × c × (change in temperature)

Using above expression, the heat lost when 200 g of water at 500 C cools to T0 C is

= 200 × 4.2 × (50 – T) = 42000 – 840T

Similarly, the heat gained when 40 g of ice at 00 C converts into water at 00 C is

= 40 × 336 J = 13440 J

In the same manner, the heat gained when temperature of 40 g of water at 00 C rises to T0 C becomes

= 40 × 4.2 × (T – 0) = 168T

We know that, the amount of heat gained is equal to the amount of heat energy lost. So, we have

=> 13440 + 168T = 42000 – 840T

=> 168T + 840T = 42000 – 13440

Or, 1008T = 28560

T = 28560 / 1008

Therefore, the final temperature T = 28.330 C

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