91Ó°ÊÓ

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.

Given,

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

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

Density is role="math" localid="1652064809462" $10.5g$

Calculating the mass = density$×$volume

$⇒10×10.5=105g$

Temperature change role="math" localid="1652064870615" $\mathrm{Δ}T={25}^0\mathrm{C}−{15}^0\mathrm{C}={10}^{\mathrm{o}}\mathrm{C}$

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

$⇒105×10×0.235=247J$

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.

Given,

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

Specific heat =role="math" localid="1652065160103" $c=0.385\mathrm{J}/{\mathrm{g}}^0\mathrm{C}$

Density is$8.92\mathrm{g}$

Calculating the mass = density$×$ volume

$⇒10×8.92=89.2g$

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

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

$⇒89.2×10×0.385=343J$