Question

A 1.00 kg block of aluminum is heated at atmospheric pressure so that its temperature increases from 22.0°C to 40.0°C.

(a) Find the work done on the aluminum._________ mJ(b) Find the energy added to it by heat.__________ kJ(c) Find the change in its internal energy.___________ kJ

Answer 1

(a). The work done on the aluminum is -48.6 mJ.

(b). The energy added to it by heat is 16.58 kJ.

(c). The change in its internal energy is 1.658 kJ.

Step-by-step explanation:

Given that,

Mass of block = 1.00 kg

Initial pressure =22.0°C

Final pressure = 40.0°C

(a). We need to calculate the work done on the aluminum

Using formula of work done

`W=-P\Delta V`

`W=-P((3\alpha m(T_(f)-T_(i)))/(\rho))`

Here,
`\Delta V=(3\alpha m(T_(f)-T_(i)))/(\rho)`

Put the value into the formula

`W=-1.01325*10^(5)*((3*24*10^(-6)*1.00(40-22))/(2700))`

`W=-48.6*10^(-3)\ J`

`W=-48.6\ mJ`

(b). We need to calculate the energy added to it by heat

Using formula of heat

`Q=mc\Delta T`

Put the value into the formula

`Q=1.0*921*(40-22)`

`Q=16.58\ kJ`

(c). We need to calculate the change in its internal energy

Using formula of internal energy

`Q=\Delta U+W`

`\Delta U=Q-W`

Put the value into the formula

`\Delta U=1658+0.0486`

`\Delta U=1658.04\ J`

`\Delta U=1.658\ kJ`

Hence, (a). The work done on the aluminum is -48.6 mJ.

(b). The energy added to it by heat is 16.58 kJ.

(c). The change in its internal energy is 1.658 kJ.