Question

7. 10 Kg of water at 90°C is cooled by mixing 20kg of water at 20°C. What

Answer 1

Answer:

Please mark me as Brilliant

Explanation:

I shall solve both one by one

Case 1: 1 kg of water at 10° C

Let the final temperature be T° C , where T°C < 30°C.

Heat give out by 4.4 kg of water at 30° C to come down to T°C = 4400 g × 1 calorie/g/°C × (30- T)° C = (4400× 30 - 4400 T) calorie

Heat gained to raise temperature of 1kg of water at 10° to T° = 1000 g × 1 calorie/g/°C × (T -10)° C = 1000 T - 10,000) calorie

Heat gained = Heat lost

1000 T - 10,000 = 1,32,000 - 4400 T; ==> 54 00T = 1.42,000; T = 1,42,000/5400 = 26.3° C.

In case it is 1 kg of water at 10° C mixed with 4.4 kg of water at 30° C, the final temperature of the mixture would be T = 26.3° C.

Case 2: 1 kg of ice.

Ice is essentially at its melting point/freezing point ie at 0° C.

Let the temperature of mixture = T° C

Heat required to melt 1 kg (=1000 g of ice) at 0° C to water at 0° C = 1000 g × 80 calorie/g = 80,000 calorie

Heat require to raise temperature of 1000 g of water at 0° C to water at T° C = 1000g × 1 calorie/g/°C× (T -0)° C = 1000 T calorie

Heat gained = 80,000 + 1000 T

Heat lost by 4.4 kg of water at 30° C to cool to T° C = 4400 g × 1 calorie/g /° C × ( 30 - T)° C = 1,32,000 - 4400 T

Heat gained = Heat lost

80,000 + 1,000 T = 1,32,000 -4400 T

5400 T = 52,000/5400; ==> T = 9.63°C.

In case it is ice, the temperature of the mixture is T = 9.63° C.

Added: around 2 pm