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# Differences between ‘heat capacity’ and ‘specific heat capacity’

• Last Updated : 02 Jun, 2022

Energy can exist in many structures and heat is one of the most interesting. Heat is regularly covered up, as it possibly exists when on the way, and is moved by various particularly various strategies. Heat move contacts each part of our lives and assists us with understanding how the universe’s capacities. It makes sense of the chill we feel on a reasonable blustery evening, or why Earth’s center still can’t seem to cool. ### Heat Capacity

Heat capacity is a physical property of matter, defined as the amount of heat to be supplied to an object to produce a unit change in its temperature. Heat capacity is also named thermal capacity.

For understanding heat capacity first and foremost we want to get heat. Heat is a sort of energy that passes from objects of higher temperature to protest of lower temperature. For instance, in the event that we contact a hot cup of espresso, we will feel hot in light of the fact that the cup moves its energy (heat) into our body. Then again, on the off chance that we contact glass of cold water, the heat energy moves from our body to the glass making it feels cold. Besides, the SI unit for heat is Joules (J). Furthermore, it is noted as Q in the situation.

Formula for Heat Capacity

Heat energy = Q/∆T

Where ‘Q’ is amount of heat

‘∆T’ specific refers to temperature

### Specific Heat Capacity

Specific heat capacity (c) of a substance is the heat capacity of an illustration of the substance divided by the mass of the model. Specific heat is also now and again insinuated as massic heat capacity. Nonchalantly, how much heat ought to be added to one unit of mass of the substance to cause an addition of one unit in temperature.

To be basic for understanding, exactly when people are out close to the sea, they have seen that the water is cold however the sand is hot. The sun is comparable there, why this differentiation in the temperature? they have probably given it a thought! The temperature of a solid and liquid part rises when we supply heat to it. Accepting that we gave a comparable proportion of the heat to two different kinds of solids then, at that point, expansion in temperature may be different in both the solids. Along these lines, dependent upon the possibility of the solid, the increase in temperature changes for different kinds of solids. This characteristic is known as Specific Heat Capacity.

Formula for Specific Heat Capacity

Specific Heat energy (c) = Q/m∆T

Where ‘Q’ is amount of heat

‘∆T’specific refers to temperature

‘m’ stands for mass

### Difference between Heat Capacity and Specific Heat Capacity

Conclusion

The main difference between heat and specific heat capacity is the difference in mass in the calculation.
Both of them are defined as the amount of energy required to raise the temperature but, in specific heat, the capacity is calculated per unit mass of a substance.

### Sample Problems

Problem 1: A piece of iron 135 g has a specific heat of 0.43 J/g°C. Also, it is heated from 100°C to 350°C. So, calculate how much heat energy is required?

Solution:

Mass (m) = 135 g

Specific heat of iron, (c) = 0.35J/g°C

Change in temperature (ΔT) = 350 – 100 = 250°C

Calculation,

Q = mc ΔT

Q = (135 g) (0.35J/g°C) (250°C)

Q = 11,812.5 J

So, the heat capacity of 135 g of iron is 11,8125.5 J.

Problem 2: It takes 427.5 J to heat 75 grams of copper from 35 °C to 65 °C. What is the specific heat in Joules/g°C?

Solution:

Mass (m) = 427.5 J

Heat of copper Q = 75 g

Change in temperature (ΔT) = 65 – 35 = 30°C

Calculation

Q = mcΔT

Putting the numbers into the equation yields:

427.5 J = (75 g)c(30°C)

Solve for c,

c = 427.5 J/(75 g)(30°C)

c = 0.19 J/g·°C

The specific heat of copper is 0.19 J/g·°C.

Problem 3: A piece of copper of 85 g has a specific heat = 0.13 J/g°C. Likewise, it is heated from 70°C to 150°C. In this way, ascertain how much heat energy is required?

Solution:

Mass (m) = 85 g

Specific heat of iron, (c) = 0.13 J/g°C

Change in temperature (ΔT) = 70 – 150 = 80°C

Calculation

Q = mc ΔT

Q = (85 g) (0.13J/g°C) (80°C)

Q = 884 J

So, the heat capacity of 85 g of copper is 884 J.

Problem 4: It takes 420.5 J to heat 55 grams of iron from 25°C to 40°C. What is the specific heat in Joules/g°C?

Solution:

Mass (m) = 420.5 g

Heat of copper Q = 55 g

Change in temperature (ΔT) = 25 – 40 = 15°C

Calculation

Q = mcΔT

Putting the numbers into the equation yields:

420.5 J = (55 g)c(15°C)

Solve for c,

c = 420.5 J/(55 g)(15°C)

c = 0.50 J/g°C

The specific heat of iron is 0.50 J/g°C.

Problem 5: If it takes 600 J of heat energy to increase the temperature of 150 g of a substance by 15˚C (without changing its phase), calculate the specific heat of the substance.

Solution:

Mass (m) = 150 g

Temperature = 15°C

Heat = 600 J

Specific heat C = q/mΔT

= 600 J/(150 g)(15°C)

= 0.266 Jg°C

Therefore Specific heat of the substance is 0.266 Jg°C

Problem 6: A 600-gram cube of lead is heated from 30°C to 80°C. How much energy was required to heat the lead? The specific heat of lead is 0.22 J/g°C.

Solution:

m = 600 grams

c = 0.22 J/g°C

ΔT = (T final – T initial) = (80°C – 30°C) = 50°C

Plug these values into the specific heat equation from above.

Q = mcΔT

Q = (600 grams)·(0.22 J/g°C) × (50°C)

Q = 6600 J

It took 6600 Joules of energy to heat the lead cube from 25°C to 75°C.

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