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If you wish to warm 120 kg of water by 16°C for your bath, how much heat is required?
((The heat capacity of liquid water is 4186 J/(kg·oC) )

Q = M C *(delta T)

delta T = 16 C (it means "change in temperature)
M = 120 kg
C = 4186 J/(kg·oC

Multiply it out for the answer (Q, the heat required), in Joules
18 years ago

Answers

Anonymous
8037120
12 years ago
cor
660
11 years ago

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