Question #293572

A body at an unknown temperature is placed in a room which is held

at a constant temperature of 32° F. If after 10 minutes the temperature of the

body is 0° F and after 20 minutes the temperature of the body is 10° F, find the

unknown initial temperature.


1
Expert's answer
2022-02-04T11:55:35-0500

Solution;

From Newton's Law of Cooling ,the Rate of cooling of the body is proportional to difference between the body's temperature and the surrounding temperature,given as;

dTdt=k(TTs)\frac{dT}{dt}=-k(T-T_s)

where T is the body's temperature and k is the proportionality constant.

Ts=32°FT_s=32°F

dTdt=k(T32)\frac{dT}{dt}=-k(T-32)

Seperate by variables;

dTT32=kdt\frac{dT}{T-32}=-kdt

Integrate;

ln(T32)=kt+Cln(T-32)=-kt+C

Rewrite;

TTs=ekt+CT-T_s=e^{-kt+C}

T=Ts+CektT=T_s+Ce^{-kt}

At t=0;

C=T0TsC=T_0-T_s

Where T0T_0 is the initial temperature.

The equation becomes;

T=Ts+(T0Ts)ektT=T_s+(T_0-T_s)e^{-kt}

Given;

t=10,T=0°F

By substitution;

0=32+(T032)e20k0=32+(T_0-32)e^{-20k}

(32T0)e10k=32(32-T_0)e^{-10k}=32

32T0=32e10k32-T_0=\frac{32}{e^{-10k}}

Also;

t=20,T=10°F

10=32+(T032)e20k10=32+(T_0-32)e^{-20k}

(32T0)e20k=22(32-T_0)e^{-20k}=22

Rewrite;

32e10ke20k=22\frac{32}{e^{-10k}}e^{-20k}=22

32e10k=2232e^{-10k}=22

10k=ln(2232)-10k=ln(\frac{22}{32})

k=0.1ln(2232)k=-0.1ln(\frac{22}{32})

Therefore;

32T0=32e10k32-T_0=\frac{32}{e^{-10k}}

32T0=320.687532-T_0=\frac{32}{0.6875}

T0=14.5°FT_0=-14.5°F



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