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

Answer:

(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.

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

$W=-P\Delta V$

$W=-P(\dfrac{3\alpha m(T_{f}-T_{i})}{\rho})$

Here,$\Delta V=\dfrac{3\alpha m(T_{f}-T_{i})}{\rho}$

Put the value into the formula

$W=-1.01325\times10^{5}\times(\dfrac{3\times24\times10^{-6}\times1.00(40-22)}{2700})$

$W=-48.6\times10^{-3}\ J$

$W=-48.6\ mJ$

(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

$Q=1.0\times921\times(40-22)$

$Q=16.58\ kJ$

(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.