2. Temperature, heat and pressure

Example problem: Cooling a cup of tea

Let’s say that we have 250mL,  of hot tea which we would like to cool down before we try to drink it. The tea is currently at 370K, and we’d like to cool it down to 350K. How much thermal energy has to be transferred from the tea to the surroundings to cool the tea?

A cup of black tea with a slice of lemon in a white teacup with a saucer.

A cup of black tea with a slice of lemon in a white teacup with a saucer.The hot tea will transfer heat to the surroundings as it cools. We are going to assume that the tea is mostly water, so we can use the density and heat capacity of water in our calculations. The specific heat capacity of water is 4.18 J/ g . K, and the density of water is  1.00g/mL.We can calculate the energy transferred in the process of cooling the tea using the following steps:

1. Calculate the mass of the substance

We can calculate the mass of the tea/water using the volume and density of water:

m = 250mL×1.00g​/mL = 250g

2. Calculate the change in temperature, ΔT

We can calculate the change in temperature, ΔT, from the initial and final temperatures:

ΔT = Tfinal​−Tinitial

= 350K – 370K

= -20K

Since the temperature of the tea is decreasing and ΔT is negative, we would expect q,  to also be negative since our system is losing thermal energy.​

3. Solve for q,

Now we can solve for the heat transferred from the hot tea using the equation for heat:

q = m × C × ΔT

= 250g × 4.18 J/g.K × -20K


Thus, we calculated that the tea will transfer 21000 J, of energy to the surroundings when it cools down from 370K to 350K.

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