An unknown material of mass 0.1 kg at a temperature of 100.0 o C is put into a 0
ID: 1621454 • Letter: A
Question
An unknown material of mass 0.1 kg at a temperature of 100.0 oC is put into a 0.150 kg Aluminum calorimeter cup that contains 0.2 kg of water. The initial temperature of the cup and water is 27.0 oC. The final temperature at equilibrium is 37.0 oC. Assuming that the calorimeter neither loses nor gains energy from the external environment, what is the specific heat of the unknown material? (Note: Specific heat of aluminum is = 910 J/kg.C; Specific heat of water is = 4180 J/kg.C)
How much heat must be added to 15.0 kg Aluminum to change it from solid at 100 oC
to a liquid at 660 oC (its melting point)?
(Heat of Fusion for Aluminum = 4.0x105 J/kg; Specific heat of Aluminum = 921 J/kg)
Explanation / Answer
Heat lost by Unknow metal = heat gained by aluminum and water
Q1 = Q2 + Q3
A = m*C*dT
Mu*Cu*dT1 = Ma*Ca*dT2 + Mw*Cw*dT3
dT1 = 100 - 37 = 63 C
dT2 = dT3 = 37 - 27 = 10 C
0.1*Cu*63 = 0.15*910*10 + 0.2*4180*10
Cu = (0.15*910*10 + 0.2*4180*10)/6.3
Cu = 1543.65 J/kg-C
B.
Heat required to melt Aluminum
Q = Q1 + Q2
Q = Ma*Ca*dT1 + Ma*La
Q = 15*921*(660 - 100) + 15*4*10^5 = 1.374*10^7 J
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