Question

An 100-g aluminum calorimeter contains 280 g of water at an equilibrium temperature of 20°C. A 170-g piece of metal, initially at 277°C, is added to the calorimeter. The final temperature at equilibrium is 32°C. Assume there is no external heat exchange. The specific heats of aluminum and water are 910 J/kg·K and 4190 J/kg·K, respectively. The specific heat of the metal is closest to:

a) 270 J/kg·K.

b) 240 J/kg·K.

c) 330 J/kg·K.

d) 390 J/kg·K.

e) 360 J/kg·K.

Answer 1

Using energy Conservation:

Energy released by metal = Energy gained by water + energy gained by Aluminum

Q1 = Q2 + Q3

Q1 = Energy released by metal = M*C*dT

M = mass of metal = 170 gm = 0.17 kg

dT = Ti - Tf = 277 - 32 = 245 C

C = Specific heat of metal = ?

Q2 = Energy released by water = Mw*CW*dT1

Mw = 280 gm = 0.28 kg

Cw = 4190 J/kg-C and dT1 = Tf - Ti = 32 - 20 = 12 C

Q3 = Energy released by aluminum = Ma*Ca*dT1

Ma = 100 gm = 0.1 kg

Ca = 910 J/kg-C and dT1 = Tf - Ti = 32 - 20 = 12 C

Now using these values:

Q1 = Q2 + Q3

0.17*C*245 = 0.28*4190*12 + 0.1*910*12

C = (0.28*4190*12 + 0.1*910*12)/(0.17*245)

C = 364.2 J/kg-K

Correct option is E.

Please Upvote.