Respuesta :

Edufirst
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

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