Water has a specific heat of 4.18 J/g°C. If 35.0 g of water at 98.8°C loses 4.94 kJ of heat, what is the final temperature of the water?
O 57.2°C
0 65.0°C
O 47.2°C
O 32.0°C

65°C

Explanation:

Formula for calorimetry is:

Q = C . m . ΔT

Where Q is heat, C is specific heat, m is mass and ΔT is the difference between final T° and initial.

We have all data to replace.

First of all, we convert kJ to J → 4.94 kJ . 1000 J / 1 kJ = 4940 Joules

We made this conversion because the unit of C, is in J

As heat is lost, we have to think in a negative value. Let's replace:

- 4940 J = 4.18 J/ g °C . 35 g . ( Final T° - 98.8°C)

- 4940 J / (4.18 J/ g °C . 35 g) =  Final T° - 98.8°C

Final T° = - 4940 J / (4.18 J/ g °C . 35 g) + 98.8°C

Final T° = 65.03 °C

Notice, that this answer has sense because as T° decreased, heat has been lost.