How much energy is needed to heat enough water to make a cup of tea (250 ml), if the water is initially at 20.0°C and you want to increase the temperature to 85.0°C? (1.00 mL of water has a mass of 1.00 g.) As the tea in the previous question steeps, it cools from 85.0°C to 75.0°C. How much energy is lost by the tea as it steeps?

Heat Energy needed (Q) = 68250 J

Energy lost (Q') = 10500 J

Explanation:

Applying,

Q = cm(t₂-t₁) Equation 1

Where Q = heat  energy needed, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

From the question,

Given: m = 250 g = 0.25 kg, t₁ = 20.0°C, t₂ = 85.0°C

Constant: c = 4200 J/kg°C

Substitute these values into equation 1

Q = 0.25(4200)(85-20)

Q = 68250 J.

Similarly, heat lost

Q' = cm(t₁-t₂) Equation 2

Where Q = heat lost

Given: t₁ = 85°C, t₂ = 75°C

Substitute into equation 2

Q' = 4200(0.25)(85-75)

Q' = 10500 J

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