What are Heat Capacity and Specific Heat? | comparison | Specific Heat Data table | numerical problems solved

Last updated on November 5th, 2021 at 04:57 am

In this post, we will discuss Heat Capacity and Specific heat, their relationship, comparison, Specific heat data tables, and formula. We will also solve a few numerical problems based on the formulas.

What is Heat Capacity or Thermal Capacity?

When an object absorbs heat, it generally results in a temperature increase of the object. (The exception to this is when a phase change occurs, such as melting ice.)

The temperature change ΔT is proportional to the heat absorbed Q.
The equation for heat transfer is Q = CΔT ………….(1)
where C is called heat capacity.

The formula of Heat capacity C = Q / ΔT ……………..(2)
 
Heat capacity depends on the type and amount of material.

Define Heat capacity or thermal capacity

The thermal capacity or heat capacity of a body is the quantity of heat needed to raise the temperature of the whole body by 1 °C.

Unit of Heat capacity or thermal capacity

Thermal capacity or Heat Capacity is measured in joules per ºC, i.e. J/ºC.

What is Specific Heat or specific heat capacity?

Specific Heat or Specific Heat Capacity c is the heat capacity per unit mass or c = C/m.

Because specific heat doesn’t depend on mass, it’s a characteristic of the type of substance itself. (The specific heat c is a property of the substance.)

In terms of specific heat capacity c (using equation 1 above), the equation for heat transfer is
Q = mcΔT ……..(3)

And the formula of specific heat: c = Q/(mΔT) …….. (4)
where m is the mass of the substance and ΔT is the change in its temperature, in units of Celsius or Kelvin. The symbol c stands for specific heat or specific heat capacity which is a property of the substance.

Define Specific Heat or specific heat capacity

The specific heat or specific heat capacity is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00 ºC.

In other words, the specific heat or specific heat capacity of a substance is the heat required to produce a 1 ºC temperature rise in 1 kg mass of that substance.

What is the SI unit of Specific Heat or specific heat capacity ?

Its SI unit is J/(kg K) or J/(kg ºC ).
The temperature change (ΔT) is the same in units of kelvins and degrees Celsius (but not degrees Fahrenheit).

What is the Specific Heat of water?

It takes 1 cal of heat to raise by 1°C the temperature of 1 g of water, so water’s specific heat is 1 cal/(g⋅°C).

Using the SI units of joule and kilogram, water’s specific heat is 4186 J/(kg⋅°C).

Heat Capacity & Specific Heat capacity – Comparison

Two objects made up of the same material but with different masses will have different heat capacities. Because Heat capacity depends on the type and amount(or mass) of material.

But, two objects made up of the same material but with different masses will have the same specific heat (or specific heat capacity). Because specific heat or specific heat capacity doesn’t depend on mass, it’s a characteristic of the type of substance itself.

Heat Capacity in terms of Specific Heat capacity – formula

When Heat Capacity (or Thermal capacity) = C
and Specific Heat Capacity (or specific heat) = c
and mass of the substance = m, then
C = m c …………….. (5)

Heat capacity = mass × specific heat capacity
C = m × c

Specific Heat Capacity data table – for solid, liquid, and gas

Substances	Specific Heat Capacity (c)
Solids	J/(kg ºC )
Aluminum	900
Asbestos	800
Concrete, granite (average)	840
Copper	387
Glass	840
Gold	129
Human body (average)	3500
Ice (average)	2090
Iron, steel	452
Lead	128
Silver	235
Wood	1700
Liquids
Benzene	1740
Ethanol	2450
Glycerin	2410
Mercury	139
Water	4186
Gases (at 1 atm constant pressure)
Air (dry)	1015
Ammonia	2190
Carbon dioxide	833
Nitrogen	1040
Oxygen	913
Steam	2020

Numerical problems based on Specific Heat Capacity(c) & Heat Capacity(C) – all solved

Problem 1:
How much energy is transferred when 50 g of brass at 20°C is heated to a temperature of 150°C? [specific heat of brass = 0.380 J/g · °C]

Solution

Step 1. The goal is to find the amount of heat transferred in joules (ΔQ). Brass has a particular specific heat and so we need to get that from the data provided or from any table available. We are assuming that the brass is not melting.

Step 2. Substitute in the given values with units: ΔQ = mcΔT = (50 g)(0.380 J/g · °C)(130°C) = 2470.0 J

Problem 2:
In an experiment, a student wishes to determine the specific heat of an unknown substance. It is observed that 100 g of the substance absorbs 3000 J of energy as it is heated from 30°C to 250°C with no change of state. What is the value of the specific heat of this material?

Solution

Step 1. To find the specific heat of the material we use our definition equation: c = ΔQ/mΔT

Step 2. Substitute in the given values with units:
c = (3000 J)/(100 g)(220°C) = 0.136 J/g · °C (very close to tungsten)