Question

Steam at 100°C is condensed into
a 38.0 g copper calorimeter cup containing 260 g of water at 27.0°C. Determine the amount of steam (in
g) needed for the system to reach a final temperature of
56.0°C. The specific heat of copper is
387 J/(kg
· °C).

Answer #1

Mass of steam = m_{s}

Mass of copper calorimeter = m_{c} = 38 g = 0.038 kg

Mass of water in calorimeter = m_{w} = 260 g = 0.26
kg

Initial temperature of calorimeter = T_{1} = 27
^{o}C

Final temperature of calorimeter = T_{2} = 56
^{o}C

Initial temperature of steam = T_{3} = 100
^{o}C

Specific heat of copper = C_{c} = 387
J/(kg.^{o}C) = 0.387 kJ/(kg.^{o}C)

Specific heat of water = C_{w} = 4.18
kJ/(kg.^{o}C)

Latent heat of vaporization of water = L = 2260 kJ/kg

m_{s}L + m_{s}C_{w}(T_{3} -
T_{2}) = m_{c}C_{c}(T_{2} -
T_{1}) + m_{w}C_{w}(T_{2} -
T_{1})

m_{s}[2260 + 4.18(100 - 56)] = (0.038)(0.387)(56 - 27) +
(0.26)(4.18)(56 - 27)

2443.92m_{s} = 31.943

m_{s} = 0.01307 kg

m_{s} = 13.07 grams

Mass of steam needed = 13.07 grams

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