Question 997: Thermodynamics & KTG

Question

A copper sphere cools from $$82^{\circ} \mathrm{C}$$ to $$50^{\circ} \mathrm{C}$$ in 10 minutes and to $$42^{\circ} \mathrm{C}$$ in the next $$10 \mathrm{~min}$$. Calculate the temperature of the surrounding?

Accepted Answer

Using Newton's Law of cooling,
$$\frac{	heta_1-	heta_2}{t}=-k\left(\frac{	heta_1+	heta_2}{2}-	heta_0ight)$$
for cooling from $$82^{\circ} \mathrm{C}$$ to $$50^{\circ} \mathrm{C}$$
$$\frac{82-50}{10}=-k\left(\frac{82+50}{2}-	heta_0ight)\quad 	ext{..... (i)}$$
for cooling from 50 to $$42^{\circ} \mathrm{C}$$
$$\frac{50-42}{10}=-k\left(\frac{50+40}{2}-	heta_0ight) \quad 	ext{..... (ii)}$$
Dividing Eq. (i) by Eq. (ii), we get
$$\frac{66-	heta_0}{46-	heta_0}=4 \Rightarrow 3 	heta_0=118 	ext { or } 	heta_0=39.3^{\circ} \mathrm{C}$$

Answer

$$18.01^{\circ} \mathrm{C}$$

Answer

$$39.3^{\circ} \mathrm{C}$$

Answer

$$10.6^{\circ} \mathrm{C}$$

Answer

$$20^{\circ} \mathrm{C}$$

Thermodynamics & KTG

A copper sphere cools from $82^{\circ} \mathrm{C} to 50^{\circ} \mathrm{C} in 10 minutes and to 42^{\circ} \mathrm{C} in the next 10 \mathrm{~min}$. Calculate the temperature of the surrounding?

Choose the correct answer: