How much heat, in joules and in calories, is required to heat 28.4g ice cube from - 23 to - 1

Given:

Mass of ice cube, m = 28.4 g

Initial temperature, T1 = - 23 C

Final temperature, T2 = - 1 C

To determine:

Heat required to raise the temperature from T1 to T2

Explanation:

Heat required (Q) is given as:

Q = mcΔT = mc (T2-T1)

where c = specific of ice (water) = 4.182 J/g C

Q = 28.4 * 4.182 * (-1 - (-23)) = 2612.9 J

Since 1 J = 0.239 cal

The corresponding Q = 0.239*2612.9 = 624.48 cal

Ans: Q = 2612.9 J

Q = 624.4 cal