If 5430 J of energy is used to heat 1.25 L of room temp. water (23.0 °C) whats the final temp of the water?

We can use the heat equation,

Q = mcΔT

Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g ⁻¹ °C⁻ ¹) and ΔT is the temperature difference (°C).

Density = mass / volume

The density of water = 0.997 g/mL

Hence mass of 1.25 L (1250 mL) of water = 0.997 g/mL x 1250 mL

= 1246.25 g

Specific heat capacity of water = 4.186 J / g °C.

Let's assume that there is no heat loss to the surrounding and the final temperature is T.

By applying the equation,

5430 J = 1246.25 g x 4.186 J / g °C x (T - 23) °C

(T - 23) °C = 5430 J / 1246.25 g x 4.186 J / g °C

(T - 23) °C = 1.04 °C

T = 1.04 °C + 23 °C

T = 24.04 °C

Hence, the final temperature of the water is 24.04 °C.