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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.

User LaTeX
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1 Answer

3 votes

Answer:

T= 25.0061 °C

Step-by-step explanation:

Given that

For iron ball:

d=5 cm

We know that for iron


C_p=0.46\ (KJ)/(Kg-K)


\rho =7870\ (Kg)/(m^3)

So the mass of iron


m=\rho (4)/(3)\pi r^3

Now by putting the values


m=7870* (4)/(3)\pi * 0.025^3

m=0.51 kg

Foe water:


C_p=4.187\ (KJ)/(Kg-K)


\rho =1000\ (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

User Avin
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