Specific Heat Capacity Formula

Specific heat of any random substance can be defined as the quantity of heat necessitated to alter the temperature of a definite quantity mass of a substance by 1 Degree.

The specific heat capacity of any random substance is the amount of energy associated with altering the temperature by a definite quantity of material or substance of 1 kg mass.

The SI unit of the specific heat capacity and specific heat is J/Kg.  

The formula of specific heat capacity can be depicted as:

Q = mcΔt

Or,

c = Q / mΔt….. (1)

Where,

Q is the heat energy 

m = mass, calculated in Kg

c Is the specific heat capacity

Δt is the change in temperature measured in kelvin. 

And the temperature change can be depicted as:

ΔT (Tf – Ti)

Where Ti Is the Initial temperature and Tf is the final temperature.

Dimension Expression of Specific Heat Capacity 

The dimensional expression of specific heat capacity can be determined as:

 the dimensional expression of heat = [M¹L²T^-2]

The dimensional expression of m = [M¹]

The dimension expression of ΔT = [K¹]

Now putting the dimensional expressions of Q, m, and ΔT, we get,

[M¹L²T^-2] / [M¹] [K¹]

After cancelling out the terms which are in common:

[M⁰ L² T^-2 K^-1]

Points to Consider 

• The transferred or displaced heat counts on three causes: the fundamental quantity or mass of the system, the state of matter of a substance, and temperature modification
• The temperature change is directly proportional to the amount of the transferred heat
• The mass or fundamental quantity is directly proportionate to the magnitude of the transferred heat
• The rate of magnitude or the amount of heat transferred depends on the phase and the material

Solved Examples

1. A portion of iron 120 g has a specific heat = 0.50 J/g∘C. Also, it is altered or heated from100∘C to 400∘C. So, determine how much heat energy is necessitated?

Mass (m) = 120 grams

Iron’s specific heat, (c) = 0.50 J/g∘C

Temperature change (ΔT) = 400 – 100 = 300 degrees Celsius

Now by using the formula of heat, we get,

Q = mc ΔT

Q = (120 g)(0.50 J/g∘C)(300 degrees Celsius)

Q = 18000 J

Hence, the capacity of the heat of 120 grams is 18000.0 J. 

2. Say that 15,245 J of heat is utilised in a copper pellet of 45 g. Also, the specific heat of copper is = 0.39J/g∘C. So, determine how much will be the alteration in temperature?  

Mass (m) = 45 g

Copper specific heat, (c) = 0.39J/g∘C.

Q = 15245 J

Now, by using the formula, we get,

Q = mc ΔT = Q / mc = ΔT

ΔT = 15245 J / (45g)(0.39Jg∘C)

ΔT = 868.66 degrees Celsius

Hence, The temperature change is about 868.66 degrees celsius.