Chemistry
posted by Raj .
A person falls in water at 0 degrees C. After 20 minutes of exposure, they are suffering acute hypothermia, as their core body temperature has dropped to 30.0 degrees C. Assume that a person’s heat capacity can be approximated by the heat capacity of water, 4.18 J/g/oC. The energy required to raise the victim’s core body temperature to 37.0 degrees C was 675 kcal. What was the mass of the victim?

q = mass x specific heat x (Tf  Ti).
q, specific heat and Tf and Ti are given. Calculate mass. Show your work if you need further assistance. Check my thinking. 
m = q/[(H)(TfTi)]
m = (4.18)[(675)(3730)]
m = 8.84 x 10^4
I'm pretty sure this is wrong. 
Wait, I did it wrong.
q = mH(Tf  Ti)
m = q/H(TfTi)
m = 675/[4.18x7]
m = 23.06
But its still the wrong answer. 
Isn't q = mass x specific heat x (TfTi)?
q = 675 kcal = 675,000 cal
mass = ??
specific heat = 4.18 J/g*c = 1 cal/g*c
Tf = 37
Ti = 30
delta T = 7
675,000 = mass x 1 x 7
mass = 675,000/7 = ?? grams.
Check my thinking. Check my work.
(That 96,000+ answer may seem large to you but that is grams and if we divide by 453.6 g/lb that converts to a little over 200 lbs which is reasonable. Watch your significant figures.) 
How did you get 1 cal/g*c....is that a calculated constant or is manual conversion required?
The answer I get using your method is 96.4kg, which is correct. 
Are we working on the same problem? I thought we were.
Your mistake in the beginning was that you didn't convert to the same set of units. You CAN'T use q in kcal and the specific heat in J/g*c. You must change q of 675 kcal to joules OR you must change the specific heat of 4.184 J/g*c to calories. It makes no difference which you change BUT one of them must be changed. I chose to change specific heat to cal and not change q. 1 calorie = 4.184 J. If you want to use 4.184 for specific heat then you must change 675 kcal by multiplying by 1000, then by 4.184.
Glad to know 96.4 kg is correct. 
energy 3 kg as heat is transfered to 0.22 kg co2 at constant pressure at an initial temperature of 30 degree estimate the final temperature of the gas