A student heated 20.0g of room temperature water in a styrofoam cup, both at 25
ID: 1836420 • Letter: A
Question
A student heated 20.0g of room temperature water in a styrofoam cup, both at 25 C by dropping into it a hot, 20g bar of copper. Copper bar was taken out of the boiling water at 100 C. Unfortunately, immediately after the first bar, a second copper bar, that cooled down to about 82 C, fell into the cup with water. What was the temperature of water with both bars in it after equilibration? Assume that the styrofoam cup absorbed 95 J of heat, while waiting for the temperature of water and the bars to equilibrate. (specific heat of copper is 0.387 J/(g C) and water 4.184 J/(g C))
Explanation / Answer
Mass of water m = 20 g
mass of copper M = 20 g
Initial temprature of water t = 25 o C
Initial temprature of copper t ' = 100 o C
Heat lost by first copper bar + heat lost by second copper bar=heat gain by water+heat absorbed by styrofoam cup
MC(100-T) + MC(82-T) = mc (T-25) + 95
Where C = Specific heat of copper = 0.387 J / g oC
c = Specific heat of water = 4.184 J / g o C
Substitute values you get ,
20(0.387)(100-T) +20(0.387)(82-T) = 20(4.184)(T-25) + 95
774-7.74 T +634.68 -7.74 T = 83.68 T - 2092 +95
83.68 T +7.74 T+7.74T = 2092 -95+774+634.68
99.16 T = 3405.68
T = 34.34 o C
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