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Question: How do you calculate the specific heat capacity of a piece of wood if \(1500gm\) of the wood absorbs...

How do you calculate the specific heat capacity of a piece of wood if 1500gm1500gm of the wood absorbs 6.75×104J6.75\times {{10}^{4}}J of heat, and its temperature changes from 32oC32{}^{o}C to 57oC57{}^{o}C?

Explanation

Solution

The specific heat of a substance is the heat required to change a unit mass of it by unit temperature. It is related to the heat absorbed or released, mass, and change in temperature, we can substitute corresponding values in the relation to calculate specific heat.
Formulas used:
c=QmΔTc=\dfrac{Q}{m\Delta T}

Complete step-by-step solution:
The specific heat of a substance is the heat required to change the temperature of 1gm1gm of the substance by 1oC1{}^{o}C. Its SI unit is J gm−1 oC−1J\,g{{m}^{-1}}\,{}^{o}{{C}^{-1}}.
c=QmΔTc=\dfrac{Q}{m\Delta T}----------- (1)
Here, cc is the specific heat
mm is the mass of the substance
ΔT\Delta T is the change in temperature
Given, for a piece of wood, mass is 1500gm1500gm, hat absorbed 6.75×104J6.75\times {{10}^{4}}J, change in temperature is-
ΔT=57−32 ⇒ΔT=15oC \begin{aligned} & \Delta T=57-32 \\\ & \Rightarrow \Delta T=15{}^{o}C \\\ \end{aligned}
We substitute the given values in eq (1) to get,
c=QmΔT ⇒c=6.75×1041500×15 ∴c=3J gm−1oC \begin{aligned} & c=\dfrac{Q}{m\Delta T} \\\ & \Rightarrow c=\dfrac{6.75\times {{10}^{4}}}{1500\times 15} \\\ & \therefore c=3J\,g{{m}^{-1}}{}^{o}C \\\ \end{aligned}
Therefore, the specific heat of wood is 3J gm−1oC3J\,g{{m}^{-1}}{}^{o}C.

Additional information: Specific heat is calculated at constant volume. At constant temperature, latent heat is calculated which changes the phase of substances at a constant temperature. The heat required to change the phase of a substance from liquid to vapor is called the latent heat of vaporization. The heat released when the state of a substance changes from liquid to solid is called the latent heat of fusion.

Note: When we have to calculate the heat absorbed or released by one mole of a substance, we calculate the molar specific heat. The specific heat is a constant and it depends on the substance and temperature; It is different for different substances and is highest for water at 4.186Jgm−1oC4.186Jg{{m}^{-1}}{}^{o}C.