Q.3.68 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the energy to heat two cubes (silver and copper), each with a volume of $10c{m}^3$from ${15}^{\circ }\mathrm{C}\text{ to }{25}^{\circ }\mathrm{C}$.

Step-by-Step Solution

Answer

the energy to heat two cubes (silver and copper) is $247J$ and $343J$ respectively.

Step 1: Given Information Part (1)

The specific heat limit of a substance is the intensity limit of an example of the substance separated by the mass of the example, additionally now and again alluded to as massic heat limit.

We have to find the energy to heat the silver cube.

Step 2: Explanation Part (1)

Given,

The volume of the silver cube = $10c{m}^3$

Specific heat = $c=0.235\mathrm{J}/{\mathrm{g}}^0\mathrm{C}$

Density is $10.5g$

Calculating the mass = density$\times$volume

$\Rightarrow 10\times 10.5=105g$

Temperature change $\mathrm{\Delta }T={25}^0\mathrm{C}-{15}^0\mathrm{C}={10}^{\mathrm{o}}\mathrm{C}$

the energy to heat the silver cube is $Q=mc\mathrm{\Delta }T$

$\Rightarrow 105\times 10\times 0.235=247J$

Step 3: Given Information Part (2)

The specific heat limit of a substance is the intensity limit of an example of the substance separated by the mass of the example, additionally now and again alluded to as massic heat limit.

We have to find the energy to heat the copper cube.

Step 4: Explanation Part (2)

Given,

The volume of the copper cube = $10c{m}^3$

Specific heat =  $c=0.385\mathrm{J}/{\mathrm{g}}^0\mathrm{C}$

Density is $8.92\mathrm{g}$

Calculating the mass = density $\times$ volume

$\Rightarrow 10\times 8.92=89.2g$

Temperature change $\mathrm{\Delta }T={25}^0\mathrm{C}-{15}^0\mathrm{C}={10}^{\mathrm{o}}\mathrm{C}$

the energy to heat the silver cube is $Q=mc\mathrm{\Delta }T$

$\Rightarrow 89.2\times 10\times 0.385=343J$