If 190 grams of water is cooled from 42.7°C to 21.2° C how much energy was lost by the water?

Energy lost by the water is 17.1 x 10³ J.

Explanation:

Specific heat is defined as the amount of energy per unit mass needed to raise the temperature of the substance by a one degree Celsius.

Heat energy gain or loss by any substance is given by:

Q = m x C x (T₁ - T₂) ... (1)

Here m is the mass of the substance, C is specific heat of the substance and T₁ and T₂ are the initial and final temperature of the substance.

According to the problem,

Mass of water, m = 190 gm

Specific heat of water, C = 4.186 J gm⁻¹ ⁰C⁻¹

Initial temperature, T₁ = 42.7⁰ C

Final temperature, T₂ = 21.2⁰ C

Substitute these values in equation (1).

Heat energy loss by water = 190 x 4.186 x (21.2 - 42.7)

= - 17.1 x 10³ J

In the above value, negative sign denotes the loss in energy.