You drop a 297-g silver figure of a polar bear into the 247-g aluminum cup of a well-insulated calorimeter containing 259 g of liquid water at 22.3°C. The bear\'s initial temperature is 98.5°C. What is the final temperature of the water, cup, and bear when they reach thermal equilibrium? The specific heats of silver, aluminum, and liquid water are, respectively, 234 J/(kg·K), 910 J/(kg·K), and 4190 J/(kg·K).
Suppose equilibrium temperature is T, then
Using energy conservation:
Heat released by Silver = Heat gained by water + aluminum
Q1 = Q2 + Q3
Ms*Cs*dT1 = Mw*Cw*dT2 + Ma*Ca*dT3
Now Using given values:
Ms = Mass of silver = 297 gm = 0.297 kg
Mw = Mass of water = 259 gm = 0.259 kg
Ma = Mass of aluminum = 247 gm = 0.247 kg
Cw = Specific heat of water = 4190 J/kg-K
Cs = Specific heat of silver = 234 J/kg-K
Ca = Specific heat of aluminum = 910 J/kg-K
dT1 = 98.5 - T
dT2 = dT3 = T - 22.3
Now Using these values:
0.297*234*(98.5 - T) = 0.247*910*(T - 22.3) + 0.259*4190*(T - 22.3)
T = (0.297*234*98.5 + 0.247*910*22.3 + 0.259*4190*22.3)/(0.297*234 + 0.247*910 + 0.259*4190)
T = 26.14 C
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