When a 83.0 g sample of a metal at 100.0 ° C is added tO₄5.0 g of water at 23.25 ° C, the final temperature of both the water and metal is 27.68 ° C. The specific heat of water is 4.184 J/g × K. What is the specific heat of the metal?

Question

Quizplus · Accepted Answer

We can use the equation:

q = mCΔT

where q is heat energy, m is mass, C is specific heat, and ΔT is the change in temperature.

For the water, we have:

q(water) = m(water)C(water)ΔT(water)
q(water) = (51.0 g)(4.184 J/g × K)(27.68 °C - 23.25 °C)
q(water) = 951 J

For the metal, we have:

q(metal) = m(metal)C(metal)ΔT(metal)
q(metal) = (83.0 g)(C(metal))(27.68 °C - 100.0 °C)
q(metal) = -5512.2 C(metal)

Notice that the ΔT for the metal is negative because it started at a higher temperature and ended up at a lower temperature.

Since the heat lost by the metal is equal to the heat gained by the water, we can set these equations equal to each other:

q(water) = -q(metal)
951 J = 5512.2 C(metal)
C(metal) = 0.139 J/g × K, which is answer choice D.

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