CGPET 2016 Solved Paper

Let the temperature of surroundings is θ0
Making use of Newton’s law of cooling,

log(64−θ0)

log(80−θ0)

=−

k

m×s

×5 ...(i)
log(

64−θ0

52−θ0

)=

k

m×s

×5
In next 10 min, the temperature becomes 52°.

log(52−θ0)

log(64−θ0)

=−

k

m×s

×10 ...(ii)
On dividing Eq. (ii) by Eq. (i), we get
log(

52−θ0

64−θ0

)=2‌log(

64−θ0

80−θ0

)
(52−θ0)×(80−θ0)=(64−θ0)2
or,
52×80−132θ0+θ02=(64)2−128θ0+θ02
4θ0=4160−4096=64
θ0=

64

4

=16°C
log(

θ−θ0

80−θ0

)=log(

64−16

80−16

)2
where,θ is the temperature of the body after 15 min.

θ−16

80−16

=

27

64

θ=43°C