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 :

Given :

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

=

= 1.83°C