Question

Rate of heat loss of a body is ‘K’ time temperature difference between body and environment. Time taken by body in losing $\frac{3}{4}rth$of the maximum heat it can lose is –

• A. $\frac{2}{K}$
• B. $\frac{2\mathcal{l}n2}{K}$
• C. $\frac{\mathcal{l}n2}{K}$
• D. $\frac{\mathcal{l}n4}{K}$

Answer 1

Answer: (B)

$\frac{2\mathcal{l}n2}{K}$

Answer 2

Let q₀ = Temperature of environment

q₁ = Initial temperature of body

q = Temperature of body at any instant t

\ Maximum heat body can lose = ms(q₁ – q₀)

[m = mass of body

s = specific heat of body]

$\frac{d\theta}{dt}$= – K(q – q₀)

$\frac{\theta - \theta_{0}}{\theta_{1} - \theta_{0}}$= e^–Kt

t = $\frac{1}{K}$ln $\left( \frac{\theta_{1} - \theta_{0}}{\theta - \theta_{0}} \right)$

Body will lose $\frac{3}{4}$ its maximum heat when

q = $\frac{\theta_{1} + 3\theta_{0}}{4}$

\ t = $\frac{2\mathcal{l}n2}{K}$