Question

A hot body placed in a surrounding of temperature θ0 obeys Newton's law of cooling dθdt=-k(θ-θ0). Its temperature at t = 0 is θ1. The specific heat capacity of the body is s and its mass is m. Find (a) the maximum heat that the body can lose and (b) the time starting from t = 0 in which it will lose 90% of this maximum heat.

Answer 1

Explanation:

According to the Newton Cooling Rule,

$\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right)$  

(a) Highest heat the body will lose, $\Delta Q_{\max }=m s\left(\theta_{1}-\theta_{0}\right)$

(b) If the body loses 90% of the maximum heat, the temperature drop will be somewhere.

$\Delta Q_{\max } \times \frac{90}{100}=m s\left(\theta_{1}-\theta\right)$

$m s\left(\theta_{1}-\theta_{o}\right) \times \frac{9}{10}=m s\left(\theta_{1}-\theta\right)$

$\theta=\theta_{1}-\left(\theta_{1}-\theta_{o}\right) \times \frac{9}{10}$

$\theta=\frac{\left(\theta_{1}-9 \theta_{0}\right)}{10}$    $\ldots e q^{n}(i)$

From Newton's Cooling Theory,

$\frac{d \theta}{d t}=-k\left(\theta_{1}-\theta\right)$

If we integrate this equation within the appropriate limit, we get

At time t = 0,

  θ = θ1

At time t,

  θ = θ  

$\int_{\theta_{1}}^{\theta} \frac{\mathrm{d} \theta}{\theta_{1}-\theta}=-k \int_{0}^{t} d t$

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

$\theta_{1}-\theta=\left(\theta_{1}-\theta_{o}\right) e^{-k t} \quad \ldots e q^{n}(i i)$

From equation (i) and equation (ii),

$\frac{\theta_{1}-9 \theta_{o}}{10}-\theta_{o}=\left(\theta_{1}-\theta_{o}\right) e^{-k t}$

$t=\ln \left(\frac{10}{k}\right)$

Answer 2

Answer:

Explanation:

Explanation:

According to the Newton Cooling Rule,

 

(a) Highest heat the body will lose,  

(b) If the body loses 90% of the maximum heat, the temperature drop will be somewhere.

   

From Newton's Cooling Theory,

If we integrate this equation within the appropriate limit, we get

At time t = 0,

 θ = θ1

At time t,

 θ = θ  

From equation (i) and equation (ii),