Question 981: Thermodynamics & KTG

Question

A cup of coffee cools from 90$$^\circ$$C to 80$$^\circ$$C in t minutes, when the room temperature is 20$$^\circ$$C. The time taken by a similar cup of coffee to cool from 80$$^\circ$$C to 60$$^\circ$$C at a room temperature same at 20$$^\circ$$C is :

Accepted Answer

$${{dT} \over {dt}} = K({T_{av}} - {T_0})$$
$${{10} \over t} = K(85 - 20)$$
$${{20} \over t} = K(70 - 20)$$
$${{t'} \over {2t}} = {{65} \over {50}}$$
$${{t'} \over t} = {{130} \over {50}}$$
$$t' = {{13} \over 5}t$$

Answer

$${5 \over {13}}$$t

Answer

$${13 \over {10}}$$t

Answer

$${13 \over {5}}$$t

Answer

$${10 \over {13}}$$t

Thermodynamics & KTG

A cup of coffee cools from 90$^\circC to 80^\circC in t minutes, when the room temperature is 20^\circC. The time taken by a similar cup of coffee to cool from 80^\circC to 60^\circC at a room temperature same at 20^\circ$C is :

Choose the correct answer: