Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 32°C such that the final temperature is 5°C. Specific heat capacity of calorimeter 0.4 J g-1 °C-1 Specific heat capacity of water = 4.2 J g-1 °C-1 Latent heat capacity of ice 330 J g-1
Topic: Heat and calorimetry
Answer
Given: water 150 g at 32 °C, calorimeter 50 g, final temperature 5 °C
c_cal = 0.4 J g⁻¹ °C⁻¹, c_w = 4.2 J g⁻¹ °C⁻¹, L = 330 J g⁻¹ Let the mass of ice be m grams, added at 0 °C.
HEAT LOST by the water and calorimeter, cooling 32 °C → 5 °C (Δθ = 27 °C):
water = 150 × 4.2 × 27 = 17 010 J calorimeter = 50 × 0.4 × 27 = 540 J total = 17 010 + 540 = 17 550 J
HEAT GAINED by the ice: melting at 0 °C = m × 330
warming 0 °C → 5 °C = m × 4.2 × 5 = 21m
total = 330m + 21m = 351m
Heat lost = heat gained: 351m = 17 550 m = 17 550 ÷ 351 m = 50 g
Mass of ice needed = 50 g
Same structure as every mixture question: everything that cools on one side, everything that warms or melts on the other.
The calorimeter counts. It is in contact with the water and cools with it, so its heat goes on the LOST side. Students who ignore the vessel get 48.5 g.
Note the latent heat here is 330, not the 336 used in other papers. Always take the value the question gives you rather than the one you remember.