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.
