Q.3.67 - An Introduction to General, Organic, and Biological Chemistry

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)

Step-by-Step Solution

Answer

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

Step 1: 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

Step 2: 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 $\mathrm{Q}=\mathrm{mc\Delta T}$

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

Step 3: 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 $Q=mc\mathrm{\Delta }T$

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