Posted by **Paula** on Tuesday, March 27, 2007 at 6:11pm.

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

- Physics -
**Anonymous**, Tuesday, November 20, 2012 at 1:23pm
8037120

- Physics -
**cor**, Thursday, March 6, 2014 at 3:20pm
660

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