A body cools from a temperature 3t to 2t in 10 min. The room temperature is t. Assume that newton's law of cooling is applicable. The temperature of the body at the end of next 10 min will be
Answers
Answer:
The temperature of the body at the end of next 10 min will be 3t/2
Explanation:
From newton's law of cooling we know that,
(θ₁ - θ₂)/t = k[(θ₁ + θ₂)/2 - θ₀]
given,
θ₁ = 3t
θ₂ = 2t
t = 10 min
θ₀ = t
hence putting these values in the above formula, we get,
(3t-2t)/10 = k[(3t + 2t)/2 - t]
=> t/10 = 3kt/2
=> k = 1/15
In the next 10 minutes, lets the temperature be θ,
applying the same formula we get,
(2t - θ)/10 = 1/15[(2t + θ)/2 - t]
(2t - θ)/10 = 1/15[θ/2]
=> 6t - 3θ = θ
=> 4θ = 6t
=> θ = 3t/2
Hence the temperature of the body at the end of next 10 min will be 3t/2
Answer:
Temperature of the body at the end of 10 min will be 3/2 times the room temperature
T' = 3/2 T
Explanation:
According to the Newton's law of cooling,
( T1 - T2) / t = K [ (T1+ T2)/2 - T ]
T1 is the initial temperature of the body.
T2 is the final temperature of the body.
T is temperature of the room.
t is the time taken to cool.
At first,
T1 = 3T
T2 = 2T
Therefore,
(3T - 2T) /10 = K [ (3T+ 2T)/2 - T ]
T/10 = K (3T/2)
=> K = 1/15
Later, on further cooling,
T1 = 2T
T2 = T'
We need to find T' in this case, which is the temperature of the body at the end of 10 minutes.
Therefore,
(2T - T') /10 = K [ (2T+ T')/2 - T ]
2T - T' = 10 * 1/15 * (T'/2)
2T - T' = T' / 3
6T - 3T' = T'
6T = 4T'
T' = 3/2 T