Question

How much heat is required to convert 10 g of ice at \( -5^{\circ} C \) into steam at \( 100^{\circ} C \). Given specific heat of ice \( 2.1 Jg -1^{\circ} C -1^{\prime \prime} \), Latent heat of steam \( =2268 J g -1 \) and latent heat of fusion of ice is \( 336 J g -1 \). Specific heat of water \( =4-2 Jg -1^{\circ} C -1 \),

Answer 1

Q₁ = mS_i ΔT = 10 × 2.1 × 5 = 105 J 

Q₂ = mL_f = 10 × 336 = 3360 J 

Q₃ = mS_w ΔT = 10 × 4.2 × 100 = 4200 J 

Q₄ = mL_v =10 × 2268 = 22680 J 

Q = Q₁ + Q₂ + Q₃ + Q₄ 

Q= 105 + 3360 + 4200 + 22680 = 30345 J

Hence, Q = 30345 J