Question

An electric heater is used to supply 54.3 J of energy to a 15.0-g graphite block with an initial temperature of 20.0 ºC. Determine the final temperature of the graphite. (Specific heat of graphite = 0.72 J/g ºC.)

Answer 1

$$Q = cm\Delta T$$

$$54.3 = 0.72*15.0*\Delta T$$

$$\Delta T = {54.3 \over 0.72*15.0} = 5.028 ^oC \approx 5.0 ^oC$$

$$T_2 = T_1+\Delta T$$

$$T_2 = 20.0+5.0 = 25.0 ^oC$$