The temperature of 170 g of water at 50°C is lowered to 5°C by adding a certain amount of ice to it. Find the mass of ice added. Given: Specific heat capacity of water = 4200 J kg-1 ⁰C-1 and Specific latent heat of ice = 336000 J kg-1
Topic: Heat and calorimetry
Answer
Given: water, mass 170 g = 0.170 kg, cooling 50 °C → 5 °C
c_w = 4200 J kg⁻¹ °C⁻¹, L = 336 000 J kg⁻¹ Let the mass of ice be m kg, added at 0 °C.
HEAT LOST by the water (Δθ = 45 °C): = 0.170 × 4200 × 45 = 32 130 J
HEAT GAINED by the ice: melting at 0 °C = m × 336 000
warming 0 °C → 5 °C = m × 4200 × 5 = 21 000m
total = 336 000m + 21 000m = 357 000m
Heat lost = heat gained: 357 000m = 32 130 m = 32 130 ÷ 357 000 m = 0.09 kg
Mass of ice added = 0.09 kg = 90 g
The constants here are per KILOGRAM, not per gram — 4200 J kg⁻¹ °C⁻¹ and 336 000 J kg⁻¹. So convert the 170 g to 0.170 kg at the start, and remember the answer will come out in kilograms and needs converting back to grams.
Mixing units is the trap: using 170 with 4200 gives an answer a thousand times too big. Look at the units printed beside each constant before you substitute anything.
There is no vessel in this question, so the lost side has only one term. The ice still has two — melt, then warm to 5 °C.