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,

first one is the correct answer

56 moles of   lieter in chemistry