5) Text (17.101) A copper calorimeter can with mass 0.446 kg contains 0.0950 kg

Question

5) Text (17.101) A copper calorimeter can with mass 0.446 kg contains 0.0950 kg of ice. The system is initially at 0.0 o C (a) if 0.0350 kg of steam at 100.0 o C and 1.00 atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents? (b) At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam? (From Table 17.3 and 17.4, water latent heat of vaporization 2.256 X 106 J/kg, water latent heat of fusion 3.34 X 105 J/kg, water specific heat 4.19 kJ/kg-K, copper specific heat 390 J/kg-K).

Explanation / Answer

Let the final temperature be T

Heat required to melt ice = 0.095 * 3.34*105 = 31730 J

Heat required to heat water + Cu to T C = 0.095 * 4.19 * 103 *T + 0.446*390*T = 571.99 T J

Heat given by steam to become water at 100 C = 0.035*2.256*106 = 78960 J

Heat lost by water at 100 C to reach T C = 0.035 * 4.19*103 * (100-T) = 146.65 (100-T)

So,

78960 + 146.65 (100 - T) = 31730 + 571.99 T

=> (571.99 + 146.65) T = 78960 + 14665 - 31730

=> T = 86.13 C

Final temperature = 86.13 C

b)

Ice and steam = 0

water = 0.095 + 0.035 = 0.130 Kg

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