Answer:
So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:
Explanation:
For this case we know the mass of the water given :
And we know that the initial temperature for this water is .
We want to cool this water to the human body temperature
Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:
Where represent the specific heat for the water and this value from tables we know that
for the water.
So then we have everything in order to replace into the formula of sensible heat and we got:
So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:
