Chemistry Forum

Hello everyone, here I am again
I have this exercise:
"A feverish student weighing 75 kg is immersed in 400 kg of water at 4.0°C to try to lower the fever. The body temperature drops from 40.0°C to 37.0°C. Considering that the specific heat of the student's body is 3.77 J/(g *°C), what is the final temperature of the water?"

I have done so:
|Q|= m*Cp*ΔT
|Q|=75000 g*3.77 J/(g*°C)*(40.0-37.0)°C=848250 J= 848.25 kJ
ΔT=|Q|/(m(H₂O)*Cp(H₂O))= 848.25 kJ/(400 kg* 1kcal/Kg*4.186kJ/kcal)=0.5°C
Tf(H₂O)=(4+0.5)°C= 4.5°C

It's correct?

Why do you mix kJ and kcal, it is the same thing only different unit.

I made the conversion of the heat capacity of the water from kcal to kJ in a single step, to multiply it with the mass of the water expressed in kg. Since I had expressed |Q|in kJ and being able to make the quotient between two measures in kJ

But the Value is given

Specific heat capacity of water is 4.19 J/(g*K)
Like the one of the human body 3.77 J/(g*K)

No need to calculate with cal.

I had seen my professor solve a similar exercise like this, but I made a mess, sorry