Question

An aluminum cup contains 225 g of water and a 40-g copper stirrer, all at 27°C. A 402-g sample of silver at an initial temperature of 86°C is placed in the water. The stirrer is used to stir the mixture gently until it reaches its final equilibrium temperature of 32°C. Calculate the mass of the aluminum cup.

Answer 1

Mass of aluminium cup = m_a

Mass of water = m_w = 225 g = 0.225 kg

Mass of copper stirrer = m_c = 40 g = 0.04 kg

Mass of silver sample = m_s = 402 g = 0.402 kg

Initial temperature of aluminium, water and copper = T₁ = 27 ^oC

Initial temperature of silver sample = T₂ = 86 ^oC

Final temperature of the mixture = T = 32 ^oC

Specific heat of aluminium = C_a = 910 J/(kg.K)

Specific heat of water = C_w = 4186 J/(kg.K)

Specific heat of copper = C_c = 390 J/(kg.K)

Specific heat of silver = C_s = 230 J/(kg.K)

The heat lost by the silver is gained by aluminium, water and copper.

m_sC_s(T₂ - T) = m_aC_a(T - T₁) + m_cC_c(T - T₁) + m_wC_w(T - T₁)

(0.402)(230)(86 - 32) = m_a(910)(32 - 27) + (0.04)(390)(32 - 27) + (0.225)(4186)(32 - 27)

m_a = 0.0452 kg

m_a = 45.2 grams

Mass of the aluminium cup = 45.2 grams