What is the change in heat energy when 75 Kg of water at 9.8°C is raised to 22.4°C?

4,000 Kilojoules (2 sig. figs. based on 2 sig. figs. in 75Kg water).    

Explanation:

Energy change equation => q = m·c·ΔT

q = total amount of heat flow

m = mass in grams = 75,000 grams

c = specific heat of water = 4.184 joules/gram·°C

ΔT = Temperature change in °C = 22.4°C - 9.8°C = 12.6°C

q =(75,000g)(4.184j/g·°C)(12.6°C) = 3,953,880 joules = 3,954 Kilojoules ≅ 4,000 Kilojoules (2 sig. figs. based on 2 sig. figs. in 75Kg water).

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