Question

\( 10 gm \) of ice at \( 0^{\circ} C \) is kept in a calorimeter of water equivalent \( 10 gm \). How much heat should be supplied to the apparatus to evaporate the water thus formed? (Neglect loss of heat) (a) \( 6200 cal \) (b) 7200 cal (c) \( 13600 cal \) (d) \( 8200 cal \)

Answer 1

Correct option is (D) 8200 cal

Heat supplied is L_fusion + McΔT + ML_vap

Q₁ = 10 × 336 + 10 × 4.2 × 100 + 10 × 2260

Q₁ = 30160 J or 7200 cal

Heat for calorimeter Q₂ = 10 × 1 × 100 = 1000 cal

Q = Q1 + Q2

= 7200 + 1000 cal

= 8200 cal