A cup of coffee cools from 90° C to 80° C in t minutes, when the room temperature is 20° C. The time taken by a similar cup of coffee to cool from 80° C to 60° C at a room temperature same at 20° C is – a. 5/13 t b. 13/10 t c. 13/5 t d. 10/13 t
By BYJU'S Exam Prep
Updated on: September 13th, 2023
The rate of loss of heat from a body is directly proportional to the difference in the temperature of the body and its surroundings Newton’s law of cooling.
From Newton’s law of cooling,
T1 – T2/ t = α (T1 + T2/ 2 – T0)
In the first case,
(90 – 80)/ t = α (90 + 80/2 – 20) …. (1)
In the second case,
(80 – 60)/t’ = α (80 + 60/ 2 – 20) …. (2)
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By solving both the equations
t’ = 13t/5
Therefore, the time taken for a similar cup of coffee to cool from 80° C to 60° C at room temperature the same at 20° C is 13t/5.
Summary:
A cup of coffee cools from 90° C to 80° C in t minutes, when the room temperature is 20° C. The time is taken for a similar cup of coffee to cool from 80° C to 60° C at a room temperature same at 20° C is –
-
5/13 t
-
13/10 t
-
13/5 t
-
10/13 t
A cup of coffee cools from 90° C to 80° C in t minutes, when the room temperature is 20° C. The time taken by a similar cup of coffee to cool from 80° C to 60° C at room temperature the same at 20° C is 13t/5.
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