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 =
T_{1} = 27 ^{o}C

Initial temperature of silver sample = T_{2} = 86
^{o}C

Final temperature of the mixture = T = 32 ^{o}C

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_{s}C_{s}(T_{2} - T) =
m_{a}C_{a}(T - T_{1}) +
m_{c}C_{c}(T - T_{1}) +
m_{w}C_{w}(T - T_{1})

(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

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