Question

Calculate the energy to heat two cubes (gold and aluminium) each with a volume of $10c{m}^3$from ${15}^{o}C$to ${25}^{o}C$. Refer to Tables 2.8 and 3.11. (3.6)

Short answer

The energy to heat two cubes (gold and aluminium) is$249J$and$242J$repectively

Step-by-step solution

introduction

the specific heat limit of a substance is the intensity limit of an example of the substance partitioned by the mass of the example, additionally in some cases alluded to as massic heat limit

explanation part (1)

Given,

the volume of gold cube= $10c{m}^3$and density = $19.3g/c{m}^3$

specific heat of gold c =$0.129\mathrm{J}/{\mathrm{g}}^0\mathrm{C}$

The mass is density$\times$volume

$10\times 19.3=193g$

The change in temperature is,

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

The heat required is $\mathrm{Q}=\mathrm{mc\Delta T}$

$193\times 0.129\times 10=249J$

explanation part (2)

Given,

the volume of aluminium cube = $10c{m}^3$and density = $2.70g/c{m}^3$

specific heat of aluminium c = $0.897\mathrm{J}/\mathrm{g}{.}^0\mathrm{C}$

The mass is density$\times$volume

$10\times 2.7=27g$

The change in temperature is,

$\mathrm{\Delta }T={25}^0\mathrm{C}-{15}^0\mathrm{C}1={10}^{\mathrm{o}}\mathrm{C}$

The heat required is $Q=mc\mathrm{\Delta }T$

$27\times 10\times 0.897=242J$