Two 0.95-kg masses,one aluminum and one tin,both at 80.0 $\circ$ C,are placed in 1.00 kg of water at 20.0 $\circ$ C.What is the equilibrium temperature of the system? The specific heats of aluminum and tin are 900 J/kg $\circ$ C and 227 J/kg $\circ$ C,respectively.
A)32.2 $\circ$ C
B)27.1 $\circ$ C
C)13.2 $\circ$ C
D)40.3 $\circ$ C

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat released by the metal as it cools down:
Q = mcΔT
For the aluminum: 
Q = (0.95 kg)(900 J/kgºC)(80ºC - T)
For the tin: 
Q = (0.95 kg)(227 J/kgºC)(80ºC - T)
Since the total amount of heat released by the metals is absorbed by the water, we can set the sum of these two equations equal to the heat absorbed by the water:
(0.95 kg)(900 J/kgºC)(80ºC - T) + (0.95 kg)(227 J/kgºC)(80ºC - T) = (1.00 kg)(4186 J/kgºC)(T - 20ºC)
Solving for T, we get T = 32.2ºC, which is approximately 32.2ºC. Therefore, the answer is (A) 32.2ºC.

Two 095-Kg Masses,one Aluminum and One Tin,both at 80 ∘\circ∘ C,are Placed in 1

Two 095-Kg Masses,one Aluminum and One Tin,both at 80 $\circ$ C,are Placed in 1