104 g of water at 30°C is taken in a calorimeter made of copper of mass 42 g. When a certain mass of ice at 0°C is added to it, the final steady temperature of the mixture after the ice has melted, was found to be 10°C. Find the mass of ice added. [Specific heat capacity of water = 4.2 J g-1 °C-1; Specific latent heat of fusion of ice = 336 J g-1; Specific heat capacity of copper = 0.4 J g-1 °C-1 ]
Topic: Principle of calorimetry
Answer
Given: water 104 g at 30 °C, copper calorimeter 42 g, final temperature 10 °C
c_w = 4.2 J g⁻¹ °C⁻¹, c_Cu = 0.4 J g⁻¹ °C⁻¹, L = 336 J g⁻¹ Let the mass of ice be m grams, added at 0 °C.
HEAT LOST by the water and calorimeter, cooling 30 °C → 10 °C (Δθ = 20 °C):
water = 104 × 4.2 × 20 = 8 736 J calorimeter = 42 × 0.4 × 20 = 336 J total = 8 736 + 336 = 9 072 J
HEAT GAINED by the ice: melting at 0 °C = m × 336
warming 0 °C → 10 °C = m × 4.2 × 10 = 42m
total = 336m + 42m = 378m
Heat lost = heat gained: 378m = 9 072 m = 9 072 ÷ 378 m = 24 g
Mass of ice added = 24 g
Set it out as two columns — everything cooling on the left, everything warming or melting on the right — and the four marks follow the four lines.
The copper calorimeter cools with the water, so it belongs on the LOST side. Leaving it out gives 23.1 g instead of 24 g; a small difference here, but the method is wrong and the marks go with the method.
The melted ice warms through 10 °C, not 20 °C, because it starts at 0 °C and ends at 10 °C. Every body in the problem has its own Δθ — write each one out rather than reusing a single number.