Question

Consider two equal 1000-cm^33 cubes of copper. Initially separated, one has a temperature of 20^\circ∘C and the other is at 100^\circ∘C. They are then brought into contact along one wall, but otherwise isolated from their surroundings. Estimate how long it will take for the two cubes to come into equilibrium.​

Answer 1

Given that,

Volume = 1000 cm³

Side = 10 cm

Temperature $T_{1}=100^{\circ}C$

Another temperature $T_{2}=20^{\circ}C$

We need to calculate the temperature T₀

Using formula of temperature for both cube

$\dfrac{kA(T_{1}-T_{0})}{a}=\dfrac{kA(T_{0}+T_{2})}{a}$

Where, a = side

$T_{1}$ = Temperature

$T_{2}$ = Another temperature

Put the value in to the formula

$T_{1}-T_{0}=T_{0}+T_{2}$

$T_{0}=\dfrac{T_{1}+T_{2}}{a}$

$T_{0}=\dfrac{100+20}{2}$

$T_{0}=60^{\circ}$

We need to calculate the heat rate

Using formula of heat rate

$H=\dfrac{kA(T_{1}-T_{0})}{a}$

Put the value into the formula

$H=\dfrac{k\times100\times10^{-4}(100-60)}{10\times10^{-2}}$

$H=4k$

We need to calculate the time take for the two cubes to come into equilibrium

Using formula of heat

$H=\dfrac{Q}{t}$

$t=\dfrac{Q}{H}$

$t=\dfrac{ms\Delta T}{4k}$

$t=\dfrac{\rho V\times s\times\Delta T}{4k}$

Put the value into the formula

$t=\dfrac{8950\times385\times10^{3}\times10^{-6}\times20}{401}$

$t=171.85\ s$

$t=2.9\ min\approx 3\ min$

Hence, It will take 3 min for the two cubes to come into equilibrium.