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Steam at 100°C is condensed into a 38.0 g copper calorimeter cup containing 260 g of...

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).
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Answer #1

Mass of steam = ms

Mass of copper calorimeter = mc = 38 g = 0.038 kg

Mass of water in calorimeter = mw = 260 g = 0.26 kg

Initial temperature of calorimeter = T1 = 27 oC

Final temperature of calorimeter = T2 = 56 oC

Initial temperature of steam = T3 = 100 oC

Specific heat of copper = Cc = 387 J/(kg.oC) = 0.387 kJ/(kg.oC)

Specific heat of water = Cw = 4.18 kJ/(kg.oC)

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

msL + msCw(T3 - T2) = mcCc(T2 - T1) + mwCw(T2 - T1)

ms[2260 + 4.18(100 - 56)] = (0.038)(0.387)(56 - 27) + (0.26)(4.18)(56 - 27)

2443.92ms = 31.943

ms = 0.01307 kg

ms = 13.07 grams

Mass of steam needed = 13.07 grams

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