If a sample of water absorbs 41.0 kJ of heat and the temperature of the sample changes from 25.0∘C to 56.0∘C

Question

If a sample of water absorbs 41.0 kJ of heat and the temperature of the sample changes from 25.0∘C to 56.0∘C, what is the mass of the sample in grams? Remember that the specific heat capacity of water is 4.184 J/g C

Answer 1

Q = mC(delta T) where Q is the amount of heat energy absorbed, m is the mass of the water, C is the specific heat capacity of the water, and delta T is the temperature rise (final temperature - initial temperature).

Therefore m = Q/(C*deltaT) = 41,000/(4.184*31) = 41000/129.704 = 316 grams