Question

Water (2750 g ) is heated until it just begins to boil. If the water absorbs 5.57×105 J of heat in the process, what was the initial temperature of the water?

Answer 1

The amount of heat absorbed,q = mcdt

Where

q = amount of heat absorbed = 5.57x10^5 J

m= mass of water = 2750 g

c = specific heat capacity of water = 4.186 J/(goC)

dt = change in temperature = final - initial

= 100 - t

Where t is the initial temperature

Plug the values we get

dt = 100-t = q/ (mc)

= 48.4

So t = 100 -48.4 = 51.6 oC

Therefore the initial temperature is 51.6oC