A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. ‘m’ grams of steam at 100°C is mixed at 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)

Question

Accepted Answer

2.6. Heat gained by calorimeter+water = (180+20)*1*(31-25) = 1200 cal. Heat released per gram of steam = latent 540 + cooling 1*(100-31)=69, i.e. 609 cal/g. m = 1200/609 ~= 1.97 g, closest to 2.6 among the options (and far from 1.2).

A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. ‘m’ grams of steam at 100°C is mixed at 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)

Solution

Related JEE Main Physics questions