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Specific Heat Calculator
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Amount of heat, required to raise the temperature of a substance by a certain amount, depends on the properties of the substance and this amount of heat varies from substance to substance. Specific heat is essentially a measure of how thermally insensitive a substance is to the addition of energy. Specific Heat Calculator will help us to calculate the specific heat of different substances.

Specific heat c of a substance is defined as the heat capacity of the substance per unit mass of the substance.

Specific Heat Formula
Specific Heat of the substance is

c = $\frac{Q}{m \Delta{T}}$

Where
c- Specific heat of the substances
Q- The energy given to the substance
m- Mass of the substances
$\Delta$T- Change in the temperature of the substance.

How to Calculate Specific Heat?

Step 1 :

Specific Heat Formula:

Specific Heat(c) = $\frac{Q}{m \Delta{T}}$

Quantity of Heat Needed(Q) = c $\times$ m $\times$ $\Delta{T}$

Mass of the Substance(m) = $\frac{Q}{c \Delta{T}}$

Change in the Temperature($\Delta{T}$) = $\frac{Q}{m \times c}$

Where
c- Specific heat of the substances
Q- The energy given to the substance
m- Mass of the substances
$\Delta$T- Change in the temperature of the substance.

Step 2 :

State the equation you plan to use and plug in the values.

Problems on Specific Heat Calculator

1. Calculate the specific heat of a substance which absorbs 4.6 joules of heat when its mass of 2.0 kg increases with respect to the change in temperature of 60$^{\circ}$C?

Step 1 :

Given that:

Quantity of Heat Needed(Q) = 4.6 joules

Mass of the substances(m) = 2.0 kg

Change in Temperature($\Delta{T}$)  = 60$^{\circ}$C

Specific Heat(c) = ?

Step 2 :

We know that:

Specific Heat(c) = $\frac{Q}{m \Delta{T}}$

Specific Heat(c) = $\frac{4.6}{2.0 \times 60}$

Specific Heat(c) = 0.0383333333 J/kg.$^{\circ}$C

Specific Heat(c) = 0.0383 J/kg.$^{\circ}$C

2. Calculate the amount of heat required for a substance of mass 3.0kg, whose specific heat is 0.632 J/kg.$^{\circ}$C during the temperature difference of 50$^{\circ}$C?

Step 1 :

Given that:

Specific Heat(c) = 0.632 J/kg.$^{\circ}$C

Mass of the substances(m) = 3.0 kg

Change in Temperature($\Delta{T}$)  = 50$^{\circ}$C

Quantity of Heat Needed(Q) = ?

Step 2 :

We know that:

Quantity of Heat Needed(Q) = c $\times$ m $\times$ $\Delta{T}$

Quantity of Heat Needed(Q) = 0.632 $\times$ 3.0 $\times$ 50

Quantity of Heat Needed(Q) = 94.8 joules