Solveeit Logo
Login

Question

Physics Question on thermal properties of matter

Certain quantity of water cools from 70∘^{\circ}C to 60∘^{\circ}C in the first 5 minutes and to 54∘^{\circ}C in the next 5 minutes. The temperature of the surroundings is

A

45∘C45 ^\circ C

B

20∘C20 ^\circ C

C

42∘C42^ \circ C

D

10∘C10 ^\circ C

Answer

45∘C45 ^\circ C

Explanation

Solution

Let TsT_s be the temperature of the
surroundings.
According to Newton's law of cooling
T1−T2t=K(T1+T22−Ts)\frac{T_1 - T_2}{t} = K \bigg( \frac{T_1+T_2}{2} - T_s \bigg)
For first 5 minutes,
T1=70∘C,T2=60∘C,t=5minutesT_1= 70 ^\circ C , T_2 = 60 ^\circ C , t=5 minutes
∴   70−605=K(70+602−Ts)\therefore \, \, \, \frac{70 - 60 }{ 5} = K \bigg( \frac{70+60}{2} - T_s \bigg)
105=K(65−Ts)\frac{10}{5} = K(65-T_s )
For next 5 minutes,
T1=60∘C,T2=54∘C,t=5minutesT_1 =60 ^\circ C , T_2 =54 ^\circ C , t= 5 minutes
∴   60−545=K(60+542−Ts)\therefore \, \, \, \frac{ 60-54}{5} = K \bigg( \frac{60+54} {2} - T_s \bigg)
65=K(57−T5)\frac{6}{5}= K(57 - T_5)
Divide eqn. (i) by eqn. (ii), we get
53=65−Ts57−Ts\frac{5}{3} = \frac{65 - T_s}{57- T_s}
285−5Ts=195−3Ts285 -5T_s =195 -3 T_s
2Ts=90   or    Ts=45∘C2T_s = 90 \, \, \, or \, \, \, \, T_s = 45 ^ \circ C