## What is Specific Heat Capacity?

**Definition of Specific Heat Capacity (SHC):** Specific heat capacity of a substance, c is defined as the heat required to raise the temperature of a unit mass of the substance by 1^{0}C or 1 Kelvin. The SI unit of specific heat capacity is in J^{0}/C or J/k, where J is Joules, C is Celsius, and K represents kelvin. We can equally measure SHC Joules per kilogram per kelvin (Jkg^{-1}K^{-1}). In this article, you will learn how to calculate specific heat capacity.

Additionally, we can equally define specific heat capacity as the quantity of heat that is needed to increase the temperature of a unit mass of a substance by one kelvin or one-degree Celsius.

**A video on How to calculate specific heat capacity by electrical method**

**The definition of** **Heat energy** is the transfer of energy from a hot substance to a cold substance due to a change in temperature. Heat energy is measured in joules.

## Heat Capacity Formula

The formula for calculating Heat energy is

H = mcΔθ

And we can also write it as

H = mc(θ_{2} – θ_{1}) because Δθ = (θ_{2} – θ_{1})

Where H = quantity of heat

m = mass of the object

Δθ = change in temperature [we measure the change in temperature in degree Celsius or kelvin]

c = specific heat capacity = constant

## Specific Heat Capacity Formula

**The formula for calculating specific heat capacity is **

c = H/mΔθ

Additionally, since the quantity of heat, H = mcΔθ

and the formula for thermal capacity, T_{c} = mc

Thus,

Quantity of heat, H = T_{c}Δθ [Where T_{c} = Thermal capacity]

and Δθ = θ_{2} – θ_{1}

H = T_{c} (θ_{2} – θ_{1})

Additionally,

Thermal capacity is measured in K^{-1}

## Table for Specific Heat Capacity of Substances

Below is a table for the specific heat capacity of substances to help understand how to calculate specific heat capacity:

Material | Specific Heat Capacity (c) |

Aluminum | 900 Jkg^{-1}K^{-1} |

Brass | 380 Jkg^{-1}K^{-1} |

Copper | 390 Jkg^{-1}K^{-1} |

Glass | 670 Jkg^{-1}K^{-1} |

Iron | 460 Jkg^{-1}K^{-1} |

Lead | 129 Jkg^{-1}K^{-1} |

Platinum | 135 Jkg^{-1}K^{-1} |

Silver | 234 Jkg^{-1}K^{-1} |

Tin | 226 Jkg^{-1}K^{-1} |

Zinc | 384 Jkg^{-1}K^{-1} |

Alcohol | 2520 Jkg^{-1}K^{-1} |

Glycerine | 2430 Jkg^{-1}K^{-1} |

Ice | 2100 Jkg^{-1}K^{-1} |

Mercury | 140 Jkg^{-1}K^{-1} |

Paraffin oil | 2130 Jkg^{-1}K^{-1} |

Turpentine | 1760 Jkg^{-1}K^{-1} |

Water | 4200 Jkg^{-1}K^{-1} |

## Solved Problems: How to Calculate Specific Heat Capacity

Here are solved problems of how to calculate specific heat capacity

### Problem 1

A copper rod with heat capacity 585JK^{-1} is heated until its temperature changes from 35^{0}C to 80^{0}C. Calculate the quantity of heat supplied to the rod. If the specific heat capacity of copper is 390 Jkg^{-1}K^{-1}, find the mass of the rod.

**Solution**

**Data:**

We start solving the problem by reading the question twice, and extracting our data while reading for the third time. This method will help us to understand the question, and apply the best formula that can help us to solve the problem.

Thermal capacity, T_{c} = 585JK^{-1}

Initial temperature, θ_{1} = 80^{0}C = 273K + 80 = 353K

Final temperature, θ_{2} = 35^{0}C = 273K + 35 = 308K

The specific heat capacity of copper, c = 390 Jkg^{-1}K^{-1}

Now we apply the formula which says:

H = T_{c} (θ_{2} – θ_{1})

After substituting our data into the above formula, we will now have:

H = 585 (353 – 308) = 585 x 45 = 26,325J = 26KJ

To find the mass of the rod, we apply the formula T_{c} = mc

Now, we substitute our data into the above formula

585 = m x 390

We can go ahead to make the m subject of the formula by dividing both sides by 390

585/390 = (m x 390)/390

And the above expression will become

m = 585/390 = 1.5 kg

Therefore, the mass of the rod is 1.5 kg

### Problem 2

What is the quantity of heat required to raise the temperature of 300 g of the aluminum cube from 30^{0}C to 70^{0}C? (Specific heat capacity of aluminum = 900Jkg^{-1}K^{-1}).

**Solution:**

**Data:**

We start solving the problem by reading the question twice, and extracting our data while reading for the third time. This method will help us to understand the question, and apply the best formula that can help us to solve the problem.

Mass of aluminium cube, m = 300 g = (300/1000 ) kg = 0.3 kg

Initial temperature, θ_{1} = 30^{0}C = 273K + 30 = 303K

Final temperature, θ_{2} = 70^{0}C = 273K + 70 = 343K

Specific heat capacity of aluminum = 900Jkg^{-1}K^{-1}

We can now apply the formula, H = mcΔθ

H = mc(θ_{2} – θ_{1}) [where Δθ = (θ_{2} – θ_{1}) ]

Apply your data to the above formula to get

H = mc(θ_{2} – θ_{1}) = 0.3 x 900 x (343 – 303) = 0.3 x 900 x 40 = 10,800J or 11KJ

Therefore, the quantity of heat required to raise the temperature of the aluminum cube is 11 – Kilojoules

### Problem 3

Calculate the change in temperature when 3000-Joules of heat is supplied to 500 grams of water

**Solution**

**Data:**

We start solving the problem by reading the question twice, and extracting our data while reading for the third time. This method will help us to understand the question, and apply the best formula that can help us to solve the problem.

Quantity of heat, H = 3000J

Mass of water, m = 500g = (500/1000)kg = 0.5kg

The specific heat capacity of water, c = 4,200 Jkg^{-1}K^{-1}

Change in temperature, Δθ =?

We can now apply the formula, H = mcΔθ and then make Δθ subject of the formula by dividing both sides by mc

H/mc = mcΔθ/mc

and mc will cancel each other on the left-hand side (LHS) leaving us with

Δθ = H/mc

Therefore, when we apply our data to the above formula, we will have

Δθ = 3000 / (0.5 x 4,200) = 3000 / 2,100 = 1.43^{0}C

and 1.43^{0}C can be converted into kelvin by performing the following operation = 273k + 1.43 = 274.43K

Therefore, the change in temperature is 1.43^{0}C (degree-celsius) or 274.43-Kelvin

### Problem 4

Calculate the quantity of heat required to raise the temperature of 5kg of copper from 25^{0}C to 100^{0}C. [The specific heat capacity of a copper = 390Jkg^{-1}K^{-1}]

**Solution**

**Data:**

Mass of the copper, m = 5kg

Initial temperature, θ_{1} = 25^{0}C = 273K + 25 = 298K

Final temperature, θ_{2} = 70^{0}C = 273K + 100 = 373K

Specific heat capacity of aluminum = 390Jkg^{-1}K^{-1}

We can now look into the formula that can help us solve the problem

H = mcΔθ

and we can now write the above formula as

H = mc(θ_{2} – θ_{1}) [where Δθ = (θ_{2} – θ_{1}) ]

By substituting our data into the above formula, we will get

H = 5 x 390 x (373 – 298) = 5 x 390 x 75 = 146,250J = 146kJ

Therefore, the quantity of heat required to raise the temperature of copper is 146 – Kilojoules.

### Problem 5

An electric heater of 60 Watts is used to heat a metal block of mass 20-kilogram for 5 minutes. Calculate the metal block’s specific heat capacity if the temperature rise is 20^{0}C.

**Solution**

**Data**

Power of the electric heater, P = 60 Watts

Mass of the metal block = 20kg

Time, t = 5 minutes = 60 x 5 = 300 seconds

Rise in temperature, Δθ = 20^{0}C

Specific heat capacity of the metal block, c =?

The formula for calculating the specific heat capacity of a substance by an electrical method is IVt = mcΔθ

And in the case of the question, we are solving now, we are dealing with the power of 60W.

Since the formula for electric Power = current x voltage = IV [ Power, P = IV]

We can now rewrite our formula IVt = mcΔθ as Pt = mcΔθ

Therefore, the right formula that fits into our data, and can help us solve the problem is Pt = mcΔθ

Now, we substitute our data into Pt = mcΔθ

Thus,

60 x 300 = 20 x c x 20

And we have

18,000 = 400 x c

we can now divide both sides by 400 to get

c = 18,000/400 = 45Jkg^{-1}K^{-1}

The specific heat capacity of the metal block, c is 45Jkg^{-1}K^{-1}

### Problem 6

A waterfall is 600m high, if the water retains 65% of the heat generated at the end of the fall, calculate the change in temperature due to the fall. [Specific heat capacity of water, c = 4,200Jkg^{-1}K^{-1}].

**Solution**

**Data:**

Height of the waterfall = 600m

Specific heat capacity of water, c = 4,200Jkg^{-1}K^{-1}

The change in temperature, Δθ =?

and we can apply the formula mgh = mcΔθ to solve the problem

Therefore, after making Δθ subject of the formula, we will have

Δθ = gh/c [ where g = gravitational acceleration = 9.8ms^{-2}]

and we can substitute our data into the above formula to get

Δθ = (9.8 x 600)/4,200 = 5,880/4,200 = 1.4^{0}C

Therefore, the change in temperature is 1.4^{0}C

### Problem 7

The hot water tap of a bath delivers water at 75^{0}C at a rate of 10 kg per minute. While a cold water tap of the birth delivers water at 25^{0}C at the rate of 22 kg per minute. If both taps are left on for 4 minutes, calculate the final temperature of the bath water, ignoring heat losses to the environment.

**Solution**

**Data:**

*For the heat of a hot water tap*

mass in 5 minutes = 10kg x 5 = 50kg

Initial temperature, θ_{1} = θ

Final temperature θ_{2} = 75^{0}C

Specific heat capacity of water, c = 4,200Jkg^{-1}K^{-1}

Therefore, the quantity of heat for hot water,** H = mc(θ _{2} – θ_{1}) = 50 x 4,200 x (75 – θ)**

*For the heat of a cold water tap*

mass in 5 minutes = 22kg x 5 = 110kg

Initial temperature, θ_{1} = 25^{0}C

Final temperature θ_{2} = θ

Specific heat capacity of water, c = 4,200Jkg^{-1}K^{-1}

Therefore, the quantity of heat for cold water,** H = mc(θ _{2} – θ_{1}) = 110 x 4,200 x (θ – 25)**

We can now see that the quantity of heat for hot water is equal to the quantity of heat for cold water

Hence, **50 x 4,200 x (75 – θ)** **= 110 x 4,200 x ( θ – 25)**

We can simplify the above expression by making** θ **subject of the formula

Therefore, our problem now becomes

**(75 – θ)** **= (110/50) x ( θ – 25)**

This is now equal to

**(75 – θ)** **= 2.2 x ( θ – 25)**

and we will have

**75 – θ** **= 2.2 θ – 55**

We can further simplify the above expression into

**75 + 55** **= 2.2 θ + θ**

Thus, **130 = 3.2 θ**

and **θ** **= 130/3.2 = 40.625 ^{0}C** =

**40.6**^{0}CTherefore, the final temperature of the water bath is **40.6 ^{0}C**

### Problem 8

An electric heater, rated 10V, 12W, fitted into a metal block supplies heat to the block of mass 1kg and a specific heat capacity of 460Jkg-1K-1. Calculate the rise in temperature in the block if the current flow for 5 minutes.

**Solution**

**Data:**

Power = 12 Watts

Time, t = 15 minutes = 5 x 60 seconds = 300s

Mass of the block = 1kg

The specific heat capacity of the metal, c = 460Jkg^{-1}K^{-1}

Rise in temperature, Δθ =?

We can now apply the formula Pt = mcΔθ to get

12 x 300 = 1 x 460 x Δθ

and we will have

3,600 = 460 x Δθ

By making Δθ subject of the formula, we will have

Δθ =3,600/460 = 7.83^{0}C

Therefore the change in temperature is **7.83 ^{0}C**

Drop a comment on any question regarding how to calculate specific heat capacity so that we can assist you.

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