Question

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?

Answer 1

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.