Question

A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. ‘m’ grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of ‘m’ is close to (Latent heat of water = 540 cal g^–1, specific heat of water = 1 cal g^–1 °C^–1) 

(1) 2 

(2) 3.2 

(3) 2.6 

(4) 4

Answer 1

(1) 2

m[540 + (100 – 31)] = 200 × [31 – 25] 

= 1200 / 609 

= 2gm