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Physics · Heat

Using specific heat capacity

You substitute into Q = mcθ but the heater time or the units never match.

The energy needed for a temperature change is Q = mcθ, where θ is the rise in temperature. If an electric heater supplies the energy, Q also equals the power times the time.

This lesson is part of heat. It follows explaining thermal equilibrium.

What do the symbols mean?

Write each symbol with its unit before you start.

Symbol Meaning Unit
Q Heat energy J
m Mass kg
c Specific heat capacity J kg⁻¹ °C⁻¹
θ Change in temperature °C

If the mass is in grams, convert it to kilograms first.

Worked example: a heater and a tank

A 500 W heater warms 2.0 kg of water from 28°C to 78°C. Use c = 4200 J kg⁻¹ °C⁻¹ and ignore heat loss. How long does it take?

  1. Temperature change θ = 78 − 28 = 50°C.
  2. Energy needed Q = 2.0 × 4200 × 50 = 420 000 J.
  3. Power is energy per second, so t = Q ÷ P = 420 000 ÷ 500 = 840 s, which is 14 minutes.

The link between the heater and the water is the energy. Both sides equal 420 000 J, which is how you check your working.

Why do real experiments give a higher c?

In a real experiment, some of the heater’s energy warms the container and the air. The water gets less than the energy supplied.

If you use the full energy supplied in Q = mcθ, the calculated c is larger than the true value. Insulating the container reduces the error.

The mistake that costs marks

The slips are in θ, in the mass units, and in the power step.

Slip Effect Fix
Using 78 as θ Energy far too large θ = 78 − 28 = 50
Mass 2000 g left as 2000 Energy 1000 times too large Convert to 2.0 kg
t = Q × P Wrong by a large factor t = Q ÷ P

Write the change in temperature as a subtraction on its own line. The units and significant figure checker can flag mismatched units.

Check yourself

A 200 W heater warms a 0.50 kg aluminium block from 30°C to 70°C. Use c = 900 J kg⁻¹ °C⁻¹ and ignore heat loss. Find the time taken.

Answer

θ = 70 − 30 = 40°C. Q = 0.50 × 900 × 40 = 18 000 J.

Time = Q ÷ P = 18 000 ÷ 200 = 90 s.

What to study next

Go on to using specific latent heat, where the temperature does not change. Then try the heat practice set.

For a teacher to work through heater questions with you, see online one-to-one Physics tuition.

Common questions

What is specific heat capacity?

It is the energy needed to raise the temperature of 1 kg of a substance by 1°C. Its unit is J kg⁻¹ °C⁻¹. A high value, like water's, means the substance absorbs a lot of energy for a small temperature rise.

Is θ the final temperature?

No. θ in Q = mcθ is the change in temperature, final minus initial. Using the final temperature alone is one of the most frequent slips in these questions.

How do I link a heater to the formula?

Energy supplied equals power multiplied by time, E = Pt. If no heat is lost, this equals Q = mcθ. Set the two equal and solve for the unknown.

Why is the measured value higher than the true one?

Some heat escapes to the surroundings and into the container. So the heater must supply more energy for the same temperature rise. Using that larger energy in the formula gives a larger c.

When heating questions combine a heater and a formula, a one-to-one Physics teacher can work through the links between power, energy and temperature with you.

  • Online one-to-one lessons for your child with an experienced teacher.
  • Your first class is a one-hour trial, from RM50. The fee is agreed before you book.
  • Happy with the teacher? Continue with lessons of about 1.5 hours. If not, ask for another teacher.