Answer
30 kcal consumed to heat water from $22^{o}C$ to $28^{o}C$
Work Step by Step
$\Delta T = 22^{o}C - 28^{o}C = -6^{o}C$
Heat = mass x $\Delta T$ x SH
= 5.0 kg x 1000g/1kg x $-6^{o}C$ x 1 cal / g $^{o}C$
= -30000 cal
1 kcal = 1000 cal
-30000 cal x 1 kcal / 1000 cal = -30 kcal
