A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at $15.0 ^ { \circ } \mathrm { C }$ In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.
A) 1900 J/kg ∙ K
B) 2270 J/kg ∙ K
C) 3300 J/kg ∙ K
D) 3800 J/kg ∙ K
E) 4280 J/kg ∙ K

Question

Quizplus · Accepted Answer

First, we need to calculate the heat released by the metal piece when it cools down from 100.0°C to the final temperature (40.0°C):
Q1 = m1c1ΔT1
where m1 is the mass of the metal piece, c1 is its specific heat (which is what we need to find), and ΔT1 is the change in temperature. 
Using the values given, we get:
Q1 = (0.4 kg)(c1)(100.0°C - 40.0°C)
Q1 = 24c1

Next, we need to calculate the heat absorbed by the water and the aluminum cup when they are heated up from 15.0°C to the final temperature (40.0°C):
Q2 = m2c2ΔT2 + m3c3ΔT2
where m2 is the mass of the water, c2 is its specific heat, m3 is the mass of the cup, c3 is the specific heat of aluminum, and ΔT2 is the change in temperature. 
Using the values given, we get:
Q2 = (0.5 kg)(4186 J/kg ∙ K)(40.0°C - 15.0°C) + (0.1 kg)(900 J/kg ∙ K)(40.0°C - 15.0°C)
Q2 = 62895 J

Finally, we know that the heat lost by the metal (Q1) is equal to the heat gained by the water and the cup (Q2) since they are in thermal equilibrium:
Q1 = Q2
24c1 = 62895 J
c1 = 2620.6 J/kg ∙ K

Therefore, the best choice is B (2270 J/kg ∙ K) since it is the closest value to 2620.6 J/kg ∙ K.

A Lab Assistant Drops a 400 15.0∘C15.0 ^ { \circ } \mathrm { C }15.0∘C

A Lab Assistant Drops a 400 $15.0 ^ { \circ } \mathrm { C }$