The temperature of a metal rod drops from 95.0 °C to 26.0 °C and loses 646.9 joules of heat. Calculate the specific heat of the metal rod that weighs 75.03 g.

Answer 1

We can use q = mC∆T to solve this problem.

q = heat = 646.9 joules

m = mass = 75.03 g

C = specific heat = ?

∆T = change in temperature = 95.0 - 26.0 = 69.0º

Solving for C, we have ...

C = q /(m)∆T) = 646.9 J / (75.03 g)(69º)

C = 0.125 J/gº