Question

If a metallic sphere gets cooled fron 62c to 50c in 10 min and in next 10 min gets cooled to 42 c then temp of surrounding

Answer 1

For this problem , we have to invoke the cooling law,

$T(t)=T_s+(T_0-T_s)e^{-kt}$ , where $T_0$ is the initial temperature of the object and $T_s$ is the temperature of the surroundings.

Since the sphere cools down from 62C to 50C in 10 minutes, we get that:

$50=T_s+(62-T_s)e^{-10k}....(1).$

We are also given that  the object cools down from 50C to 42C in the next ten minutes, we get that,

$42=T_s+(62-T_s)e^{-20k}....(2).$

Dividing equation (1) by (2), we get that ,

$\frac{50}{40}= \frac{T_s+(62-Ts)e^{e^{-10k}}}{T_s+(62-T_s)e^{-20k}} \\=>\frac{50}{42}=e^{-10k-(-20k)} =e^{10k}\\=>k=\frac{\ln({\frac{50}{42})}}{10} =0.0174$

The equation for the cooling of this object is,

$T(t)=T_s+(T_0-T_s)e^{-0.0174t}.$

Using equation (1) we get that ,

$50=T_s+(62-T_s)e^{-10(0.0174)}$

From this it follows that,

$50=T_S+(62-T_s)\times 0.84.$

Solving for Ts in this equation we get that

$-12=T_S-0.84T_S.$

From this information, we get that $T_s=-75$.

The temperature of the surroundings in this problem is -75C.

Answer 2

Answer:

Explanation:

hope it's clear