# Thermal Properties Of Matter

In this article, we are going to learn about the Thermal Properties of Matter. We will cover the following subtopics in this article:

• Internal Energy
• Heat Capacity
• Specific Heat Capacity
• Heat = mass × specific heat capacity × temperature change

## Internal Energy

The Kinetic Model of Matter states that all matter is made up of particles in constant random motion.

The temperature of a substance is directly proportional to the average kinetic energy of its particles (that made up the substance).

## Heat Capacity

Heat capacity is the amount of thermal energy required to raise the temperature of a substance by 1 K or 1ºC.

$\displaystyle{C = \frac {Q}{∆θ}}$

Where $\displaystyle{C}$ is the Heat Capacity;

$\displaystyle{Q}$ is the Thermal Energy in joules (J);

$\displaystyle{∆θ}$ is the change in temperature in (K or ºC).

The SI unit of Heat Capacity C is J K⁻¹

Another common unit is J ºC⁻¹

The factors that affect heat capacity C are:

1. The type of material
2. The amount (mass) of substance

## Specific Heat Capacity

Specific heat capacity c is the amount of thermal energy required to raise the temperature of a unit mass (e.g. 1 kg) of a substance by 1 K (or 1°C).

\begin{align*} \displaystyle c &= \frac {C}{m}\\[2ex] &= \frac {Q}{m∆θ} \end{align*}

where,

Q (J) = thermal energy required

Δθ (K or °C) = change in temperature

m (kg) = mass of substance

The SI unit of c is J kg⁻¹ K⁻¹.

Another common unit of c is J kg⁻¹ °C⁻¹.

## Heat capacity and specific heat capacity

Given that the thermal energy needed for 100 g of water to increase its temperature by 1°C is 420 J.

Heat Capacity = 420 J

Specific heat capacity

= $\displaystyle{\frac {\;C\;}{\;m\;}}$

= 420 J °C⁻¹ ÷ 100 g

= 4.2 J °C–1g–1

Or

Specific heat capacity c

= $\displaystyle{\frac {420}{0.1}}$

= 4200 J K⁻¹ kg⁻¹

Example 1:

Determine the quantity of heat required to raise the temperature of 100 g of ice from −20°C to −5°C.

The specific heat capacity of ice is 2000 J kg⁻¹ K⁻¹.

Solution:

Δθ = −5°C − (−20°C)

= 15°C

Q = mcΔθ

= 0.1 kg × 2000 J kg⁻¹ K⁻¹ × 15°C

= 3000 J

Example 2:

Determine the specific heat capacity of aluminium, given that a heater rated 30 W takes 5 min to raise the temperature of 450 g of aluminium from 27°C to 50°C.

Solution:

Δθ = 50°C − 27°C

= 23°C

Work done, Energy = Power × Time

P × t   =    mcΔθ

30 W × 5 × 60 s   =   0.45 kg × c × 23°C

c   =    870 J kg⁻¹ °C

Example 3:

A piece of hot coal of mass 50 g is at a temperature of 200°C. It is dropped into 150 g of water at a temperature of 25°C. Determine the final temperature reached, assuming negligible heat loss to the surroundings. (The specific heat capacity of coal and water are 710 J kg⁻¹ K⁻¹ and 4200 J kg⁻¹ K⁻¹ respectively.)

Solution:

Thermal energy is transferred from the hot coal to the cooler water.

Let the final temperature be θ.

By the principle of conservation of energy,

Thermal energy lost by coal = Thermal energy gained by water

mcccΔθ =  mwcwΔθw

0.05 × 710 × (200 − θ)   =  0.15 × 4200 × (θ − 25)

7100 − 35.5θ   =  630θ − 15 750

594.5θ   =  22 850

θ   =  38.4°C

## Practice Questions

Question 1:

A heater rated 25 W takes 3 min to raise the temperature of 65 g of aluminium from - 27°C to - 1°C. Calculate the specific heat capacity of ice. So,

Solution:

Δθ = - 1°C − (− 27)°C

= 26°C

So,

P × t   =    mcΔθ

25 W × 3 × 60 s   =   0.065 kg × c × 26°C

c   =    2663 J kg⁻¹ °C⁻¹

Question 2:

Some steel balls of total mass of 150g at a temperature of 0°C are dropped into 200g of boiling water at a temperature of 100°C.

(The specific heat capacity of steel is 420 J kg$^{-1}$°C​$^{-1}$ and water is 4200 J kg$^{-1}$°C$^{-1}$.)

What is the final temperature of a mixture at equilibrium?

Solution:

Thermal energy is transferred from the boiling water to the steel balls.

Let the final temperature be $\theta$.

By the Principle of Conservation of Energy,

Thermal energy lost by water = thermal energy gained by steel balls.

$m_wC_w\Delta \theta_w=m_sC_s\Delta \theta_s$

$0.2 \times 4200 \times (100 - \theta)=0.15 \times 420 \times (\theta -0)$

$84000 - 840 \theta=63\theta$

$903\theta=84000$

$\theta=93.0$°C.

Example 4:

It takes the same heater 500 s to heat up a piece of iron by 5 K, and 200 s to heat up a piece of copper by 10 K. Which piece of metal has the larger mass?

Assume no heat loss and that the specific heat capacity of iron and copper are 460 J kg⁻¹ K⁻¹ and 400 J kg⁻¹ K⁻¹ respectively.

Solution:

The pieces of copper and iron have different masses and undergo different temperature changes, but the same heater is used, so P is constant and common to both.

By the Principle of Conservation of Energy,

electrical energy supplied   =   heat gained

P × 200   =   mcopper × 400 × 10   ------------ (1)

P × 500   =   miron × 460 × 5         ------------ (2)

(1) ÷ (2), 0.4   =   (mcopper ÷ miron) × 1.74

Re-arranging,                                           miron    =   4.35 mcopper

Therefore, the mass of the piece of iron is 4.35 times that of the piece of copper.

## Conclusion

In this article, we have discussed the Thermal Properties of Matter. We have learnt about:

• Internal Energy
• Heat Capacity
• Specific Heat Capacity
• Heat = mass × specific heat capacity × temperature change
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