Question

If 234 mL of water at 35*C loses 1412 calories of heat, what is the new temperature of the water following this heat loss?

Answer 1

Heat loss, Q = 1412 cal
mass of water, V x d = 234 ml x 1 g/ml = 234 g
Specific heat, C = 1 cal/g °C

Q = mCΔT ⇒ΔT = Q / mC

To - Tf = Q / mC
Tf = To - Q / mC

Tf = 35° - 1412 cal/[234 g * 1cal/g°C]

Tf = 35° - 6.03°

Tf = 28.97 °C