Question

A 5 cm diameter iron ball at 80°C is dropped into an insulated tank that contains 0.5 m^3 of liquid water at 25°C. Determine the temperature when thermal equilibrium is reached.

Answer 1

Answer:

T= 25.0061 °C

Explanation:

Given that

For iron ball:

 d=5 cm

We know that for iron

$C_p=0.46\ \frac{KJ}{Kg-K}$

$\rho =7870\ \frac{Kg}{m^3}$

So the mass of iron

$m=\rho \dfrac{4}{3}\pi r^3$

Now by putting the values

$m=7870\times \dfrac{4}{3}\pi \times 0.025^3$

m=0.51 kg

Foe water:

$C_p=4.187\ \frac{KJ}{Kg-K}$

$\rho =1000\ \frac{Kg}{m^3}$

m=1000 x 0.5 kg

m=500 kg

So

heat loss by iron ball = heat gain by water

$\left (mC_p\Delta T\right )_{iron}=\left (mC_p\Delta T\right )_{water}$

Lets take T is the final temperature

0.51 x 0.46 (80 -T) = 500 x 4.187 x (T-25)

T= 25.0061 °C