Question

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).

Answer #1

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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