# a) A body cools down from 50 deg C to 45 deg C in 5 minutes and to 40 deg C in another 8 minutes....

## Question:

a) A body cools down from 50°C to 45°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surrounding.

b) A calorimeter contains 50 g of water at 50°C. The temperature falls to 45°C in 10 minutes. When the calorimeter contains 100 g of water at 50°C, it takes 18 minutes for the temperature to become 45°C. Find the water equivalent of the calorimeter.

## Newton's law of cooling:

Newton's law of cooling states that, as the subtraction of temperature between the object and surrounding increases, the rate of heat change increases. When the transfer of heat takes place through radiation, change in temperature is minimum.

(a)

Given data

• The initial temperature of body is {eq}{T_1} = {50^{\rm{o}}}{\rm{C}} {/eq}.
• The temperature of body is {eq}{T_2} = {45^{\rm{o}}}{\rm{C}} {/eq}.
• The time taken to fall in temperature is {eq}{t_1} = 5\;\min {/eq}.
• The final temperature of body is {eq}{T_3} = {40^{\rm{o}}}{\rm{C}} {/eq}.
• The time taken to fall in temperature is {eq}{t_2} = 8\;\min {/eq}.

The expression for the newton?s law of cooling is given as,

{eq}\dfrac{{dT}}{{dt}} = - K\left[ {{T_\infty } - {T_s}} \right]......\left( 1 \right) {/eq}

In second case,

The average temperature is,

{eq}\begin{align*} {T_{av,2}} &= \dfrac{{45 + 40}}{2}\\ {T_{av,2}} &= {42.5^{\rm{o}}}{\rm{C}} \end{align*} {/eq}

The difference in the temperature of body and surrounding is,

{eq}\Delta T = {\left( {42.5 - {T_o}} \right)^{\rm{o}}}{\rm{C}} {/eq}

The rate of fall of temperature is,

{eq}\begin{align*} \dfrac{{dT}}{{dt}} &= \dfrac{{\Delta T}}{t}\\ \dfrac{{dT}}{{dt}} &= \dfrac{{45 - 40}}{8}\\ \dfrac{{dT}}{{dt}} &= {\dfrac{5}{8}^{\rm{o}}}{\rm{C/min}} \end{align*} {/eq}

Substituting the values in equation 1,

{eq}\dfrac{{dT}}{{dt}} = - K\left( {42.5 - {T_o}} \right)......\left( 2 \right) {/eq}

In Case 1,

The average temperature is,

{eq}\begin{align*} {T_{av}} &= \dfrac{{50 + 45}}{2}\\ {T_{av}} &= {47.5^{\rm{o}}}{\rm{C}} \end{align*} {/eq}

The difference in the temperature of body and surrounding is,

{eq}\Delta T = {\left( {47.5 - {T_o}} \right)^{\rm{o}}}{\rm{C}} {/eq}

The rate of fall of temperature is,

{eq}\begin{align*} \dfrac{{dT}}{{dt}} &= \dfrac{{\Delta T}}{t}\\ \dfrac{{dT}}{{dt}} &= \dfrac{{50 - 45}}{5}\\ \dfrac{{dT}}{{dt}} &= {1^{\rm{o}}}{\rm{C/min}} \end{align*} {/eq}

Substituting the values in equation 1,

{eq}\dfrac{{dT}}{{dt}} = - K\left( {47.5 - {T_o}} \right)......\left( 3 \right) {/eq}

Divining equation 2 and 3,

{eq}\begin{align*} \dfrac{1}{{\dfrac{5}{8}}} &= \dfrac{{47.5 - {T_o}}}{{42.5 - {T_o}}}\\ 0.625\left( {47.5 - {T_o}} \right) &= 42.5 - {T_o}\\ {T_o} &= {34.1^{\rm{o}}}{\rm{C}} \end{align*} {/eq}

Thus, the temperature of the surrounding is {eq}{34.1^{\rm{o}}}{\rm{C}} {/eq}.

(b)

Given data

• The mass of water in calorimeter is {eq}{m_1} = 50\;{\rm{g}} {/eq}.
• The temperature of the water is {eq}{t_{w,i}} = {50^{\rm{o}}}{\rm{C}} {/eq}.
• The temperature of water decreases to {eq}{t_{w,f}} = {45^{\rm{o}}}{\rm{C}} {/eq}.
• The time taken for the change in temperature is {eq}{t_1} = 10\;{\rm{min}} {/eq}.
• The mass of water in calorimeter is {eq}{m_2} = 100\;{\rm{g}} {/eq}.
• The time taken for the change in temperature is {eq}{t_2} = 18\;{\rm{min}} {/eq}.

The expression for the rate of heat flow is given as,

{eq}q = \dfrac{{ms\Delta t}}{t}......\left( 1 \right) {/eq}

Substituting the values in equation 1, for rate of heat flow,

{eq}{q_1} = \dfrac{{\left( {w + 50 \times {{10}^{ - 3}}} \right) \times 4200 \times 5}}{{10}}......\left( 2 \right) {/eq}

Substituting the values in equation 1, for rate of heat flow,

{eq}{q_2} = \dfrac{{\left( {w + 100 \times {{10}^{ - 3}}} \right) \times 4200 \times 5}}{{18}}......\left( 3 \right) {/eq}

Equating equation 2 and 3,

{eq}\begin{align*} \dfrac{{\left( {w + 50 \times {{10}^{ - 3}}} \right) \times 4200 \times 5}}{{10}} &= \dfrac{{\left( {w + 100 \times {{10}^{ - 3}}} \right) \times 4200 \times 5}}{{18}}\\ 18\left( {w + 50 \times {{10}^{ - 3}}} \right) &= 10\left( {w + 100 \times {{10}^{ - 3}}} \right)\\ 18w - 10w &= 1 - 0.9\\ w &= 12.5 \times {10^{ - 3}}\;{\rm{kg}} \end{align*} {/eq}

Thus, the water equivalent of the calorimeter is {eq}12.5\;{\rm{g}} {/eq}.