Question

1. A 25.35-g piece of iron absorbs 1562.75 Joules of heat energy, and its temperature
changes from 30°C to 185°C. Calculate the specific heat capacity of iron.

Answer 1

Answer:

Q = C M T        where C is the specific, M the mass, T the temperature change

Note 1 cal = 4.19 Joules

1562.75 J / (4.19 J/cal) = 378 cal

C = Q / (M * T) = 378 cal / (25.35 g * 155 deg C)

C = .096 cal / g deg C