Curd is at 80° F, five minutes later it came down at 60°F. After another 5 minutes, its temperature became 50°F. Given that the rate of change of temperature is proportional to (T - S), where S is tempreature of surroundings and T is temperature of the curd at any time t. Then the temperature of the surroundings is:

40°F

The correct answer is Option (2) → 40°F

According to Newton's law of cooling:

$\frac{dT}{dt}=-k(T-S)$

Solution: $T(t)=S+(T_0-S)e^{-kt}$

At $t=0,\;T_0=80$

At $t=5,\;T=60 \;\;\Rightarrow\;\;60=S+(80-S)e^{-5k}$

At $t=10,\;T=50 \;\;\Rightarrow\;\;50=S+(80-S)e^{-10k}$

From first condition:

$60-S=(80-S)e^{-5k}$ …(1)

From second condition:

$50-S=(80-S)e^{-10k}$ …(2)

Divide (2) by (1):

$\frac{50-S}{60-S}=\frac{(80-S)e^{-10k}}{(80-S)e^{-5k}}=e^{-5k}$

So, $e^{-5k}=\frac{50-S}{60-S}$ …(3)

From (1): $\;60-S=(80-S)e^{-5k}$

Substitute (3):

$60-S=(80-S)\cdot\frac{50-S}{60-S}$

$(60-S)^2=(80-S)(50-S)$

$3600-120S+S^2=4000-130S+S^2$

$-120S+3600=-130S+4000$

$10S=400$

$S=40$

Temperature of surroundings = $40^\circ F$