If you add 6.20 kJ of heat to a 48.80 gram piece of metal originally at room temperature (25°C), how hot will this piece of metal become? The specific heat of the metal is 0.390 J/g°C.

Question

If you add 6.20 kJ of heat to a 48.80 gram piece of metal originally at room temperature (25°C), how hot will this piece of metal become? The specific heat of the metal is 0.390 J/g°C.

Answer 1

q = mC∆T

q = heat = 6.20 kJ

m = mass = 48.80 g

C = specific heat = 0.390 J/g/deg

∆T = change in temperature

Before moving on, we must either change the heat (q) to J, or change the specific heat (C) to units of kJ/g/deg, so that we have the same units. I'll change the units of q to 6200 J from 6.20 kJ

q = mC∆T

6200 J = (48.80 g)(0.390 J/g/deg)(∆T)

∆T = 6200 J/(48.80 g)(0.390 J/g/deg) = 326 degrees

Final temperature of the metal = 25º + 326º = 351ºC