Steam at 100° C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be: [Take specific heat of water = 1 cal g1 °C-1 and latent heat of steam = 540 cal g-1]
Answer & explanation
Correct answer: option 4
According to the principle of calorimetry
Heat lost = Heat gained
mLvaporization + msw\(\Delta T\) = mwsw\(\Delta T\)
m×540 + m×1×(100-80)= 20×1×(80-10)
m = 2.5 g
Total mass of water= 20 g + 2.5 g = 22.5 g