KVPY 2019 SB SX Exam Previous Papers

Question : 99 of 120

Marks: +1, -0

The circuit below is used to heat water kept in a bucket.

Assume heat loss only by Newton's law of cooling, the variation in the temperature of the water in the bucket as a function of time is depicted by :

Solution:

Rate of heat gained by water

ms[{dT}/{dt}]=i^2R_1-4σ eAT^3_0[T-T_0]

{dT}/{dt}={i^2R_1}/{ms}-{4σ eAT^3}/{ms}(T-T_0)

{dT}/{dt}=C_1-C_2

(here

C_1

and

C_2

are positive constant)

∫↖{T}↙{T_1}{dT}/{C_1-C_2T}=∫dt

1/{-C_2}ln{C_1-C_2T}/{C_1-C_2T_i}=t

C_1 -C_2 T = (C_1 -C_2T_i)e^{-c^2 t }

C_2 T = C_1 -(C_1 -C_2 T_i) e^{-c^2 t }

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