Kilocalories to heat 5.0 kg of water from 22°C to 28°C

Answer 1

q = mC∆T

q = heat = ?

m = mass = 5.0 kg

C = specific heat of water = 4.184 J/gº = 4.184 kJ/kgº

∆T = change in temperature = 6º

q = (5.0 kg)(4.184 kJ/kgº)(6º)

q = 125.5 kJ = 130 kJ (2 sig. figs.)