A 45-kg sample of water absorbs 345 kJ of heat. If the water was initially at 22.1 °C, what is its final temperature?
Problem 82P :
Sample of amount of water = 45 kg
Heat absorbed = 345 kJ
Initial temperature of water = 22.1 °C.
Final temperature = ?
To calculate the final temperature, we need specific heat of water :
The Specific Heat of Iron = 4.186 J/g/°C (K).
Now, let’s calculate the change in temperature () :
Q(Amount of heat) = m (mass) C(Specific heat) (Temperature change)
345 kJ = 45 kg 4.186 J/g/°C
345 kJ = 188.37