Question

Consider two equal 1000-cm3 cubes of copper. Initially separated, one has a temperature of 20 C and the other is at 100C. 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

As per the question, we have;

Volume = $1000 cm^3$

Side = $10 cm$

Temperature 1 = $T_1$ = $100$°C

Temperature 2 = $T_2$ = $20$°C

Now, In order to find the temperature $T_0$, we use this formula;

$\frac{kA(T_1-T_0)}{a} = \frac{kA(T_0+T_2)}{a}$

in which,

a = side of cube

$T_1$ = Temperature 1

$T_2$ = Temperature 2

So,

$T_0 = \frac{T_1 + T_2}{a}$

⇒ $T_0 = (\frac{100 + 20}{2} )$ °C = $60$°C

Now, to find the heat rate;

$H = \frac{kA (T_1-T_0)}{a}$ $k$ = $\frac{k * 100 * 10^-^4 (100-60)}{10*10^-^2}$ $k$ = $4k$

Now, the formula of heat is used in order to to derive the time take for the two cubes to come into equilibrium;

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

⇒ $t = \frac{Q}{H}$ = (Δ$T$) ÷ $4k$ = (ρ$V$ × $s$ × Δ$T$) ÷ $4k$

⇒ $t =$ $\frac{8950 * 385 * 10^3 * 10^-^6 * 20}{401}$ secs = 171.85 secs = [2.9] mins = 3 mins

Ans) It will take approximately 3 minutes for the two cubes to come into equilibrium

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