In this post, we will solve a few interesting numerical problems based on specific heat capacity. Learn more about Specific heat capacity here.

## Solving numerical problems using specific heat capacity formula

**Question 1]** A tank holding 30 kg of water is heated by a 3 kW electric immersion heater. If the specific heat capacity of water is 4200 J/(kg ºC), estimate the time for the temperature to rise from 10 ºC to 60 ºC.

**Solution**:

A 3 kW (3000 W) heater supplies 3000 J of heat energy per second.

Let t = time taken in seconds to raise the temperature of the water by (60−10) = 50 ºC,

∴ heat supplied to water in time t seconds = (3000 × t) J …………. (1)

From the heat equation, we can say heat received by water =mass x specific heat capacity x temperature rise

= 30 kg × 4200 J/(kg ºC) × 50 ºC = 30 x 4200 x 50 J…………….. (2)

**Assuming heat supplied = heat received,**

Hence, from equations 1 & 2 we get:

3000 × t = (30 × 4200 × 50)

t = (30 × 4200 × 50) / 3000 = 2100 seconds = 35 minutes

**So the time estimate is 35 minutes. (answer)**

**Question 2**]

A piece of aluminium of mass 0.5 kg is heated to 100 ºC and then placed in 0.4 kg of water at 10 ºC. If the resulting temperature of the mixture is 30 ºC, what is the specific heat capacity of aluminium if that of water is 4200 J/(kg ºC)?

**Solution:**

When two substances at different temperatures are mixed, heat flows from the one at the higher temperature to the one at the lower temperature until both are at the same temperature i.e., the temperature of the mixture.

If there is no loss of heat, then in this case:

Heat given out by aluminium = Heat taken in by water*Using the heat equation and letting c be the specific heat capacity of aluminium in J/(kg ºC), we have*

heat given out = 0.5 kg × c × (100 − 30) ºC

heat taken in = 0.4 kg × 4200 J/(kg ºC) × (30 − 10) ºC

∴ 0.5 kg × c × 70 ºC = 0.4 kg × 4200 J/(kg ºC) × 20 ºC

35 c = 8 x 4200

c = ( 8 x 4200)/35 J/(kg ºC) = 960 J/(kg ºC)

Hence, the specific heat capacity of aluminium 960 J/(kg º