A cast iron grate is used to sear "grill marks" into food. If the grate has a heat capacity of 458.5 J/K, how much heat must be added to the grate to heat it from 292.5K to 459.7K?

2.74 Joules

Explanation:

The heat capacity of the metal is the amount of heat needed to raise the temperature of the metal by 1 k

heat capacity c of the sear grill = 458.5 J/K

initial temperature T1  = 292.5 K

final temperature T2 = 459.7 K

temperature change ΔT= 459.7 - 292.5 = 167.2 K

since the heat capacity is heat energy needed per degree rise in Kelvin, then the heat to raise the metal by this temperature change will be

heat required = c/ΔT = 458.5/167.2 = 2.74 Joules