Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

A system $S$ receives heat continuously from an electrical heater of power $10W$. The temperature of $S$ becomes constant at ${50^ \circ }C$ when the surrounding temperature is ${20^ \circ }C$. After the heater is switched off $S$ cools from ${35.1^ \circ }C$ to ${34.9^ \circ }C$ in \[1\] minute. The heat capacity of $S$ is:
$(A)100J{/^ \circ }C$
$(B)300J{/^ \circ }C$
$(C)750J{/^ \circ }C$
$(D)1500J{/^ \circ }C$

seo-qna
Last updated date: 19th Apr 2024
Total views: 35.1k
Views today: 0.35k
Answer
VerifiedVerified
35.1k+ views
Hint: We can solve the given problem with the help of Newton’s law of cooling. It states that the transfer of heat is proportional to the temperature difference.

Formula used:
$ \Rightarrow Q = C \times \Delta t$
Where, $Q$ is the heat, $\Delta t$ is the temperature difference, $C$ is the heat capacity.

Complete step by step answer:
First, let us see Newton's law of cooling. Newton’s law of cooling states that the rate of heat loss is directly proportional to the difference in the temperature of its surroundings and the temperature of the body.
We are going to solve the whole problem using the formula,
$ \Rightarrow Q = C \times \Delta t$
Where, $Q$ is the heat, $\Delta t$ is the temperature difference,$C$ is the heat capacity.
Now we can solve the given question with the help of newton’s law of cooling. We have,
$ \Rightarrow Q \propto \left( {{T_{body}} - {T_{surrounding}}} \right)$
Where $Q$ is the heat and $T$ is the temperature.
We can differentiate the value of $Q$ with respect to time. We get,
$ \Rightarrow \dfrac{{dQ}}{{dt}} \propto \left( {{T_{body}} - {T_{surrounding}}} \right)$
At ${50^ \circ }C$ the value of $Q$ is given as,
$ \Rightarrow \dfrac{{dQ}}{{dt}} \propto \left( {50 - 20} \right)$
We can introduce the term constant $k$to remove the proportionality symbol. We get,
$ \Rightarrow \dfrac{{dQ}}{{dt}} = k\left( {50 - 20} \right)$
The whole equation has a value of 10. That is,
$ \Rightarrow \dfrac{{dQ}}{{dt}} = k\left( {50 - 20} \right) = 10$
We can simplify the given equation we get,
$ \Rightarrow k\left( {30} \right) = 10$
Therefore, the value of $K$ is $\dfrac{1}{3}$. That is,
$ \Rightarrow k = \dfrac{1}{3}$
We need to calculate the mean temperature. The value of mean temperature is,
$ \Rightarrow {T_{meam}} = \dfrac{{50 + 20}}{2}$
We divide and we get,
$ \Rightarrow {T_{meam}} = {35^ \circ }C$
We can calculate the energy loss value. The energy loss is given by the heat divided by the time. We get,
$ \Rightarrow \dfrac{{dQ}}{{dt}} = k\left( {{T_{mean}} - {T_{\min }}} \right)$
We can substitute the values in the given equation we get,
$ \Rightarrow \dfrac{{dQ}}{{dt}} = \dfrac{1}{3}\left( {{{35}^ \circ }C - {{20}^ \circ }C} \right)$
After simplifications we get,
$ \Rightarrow \dfrac{{dQ}}{{dt}} = 5J/s$
In the question, it is given that 1 minute, 1 min is equal to 60 sec. we have to multiply this time value with the energy loss value. We get,
$ \Rightarrow dQ = 5J/s \times 60$
After multiplying the values, we get,
$ \Rightarrow dQ = 300J$
Now we can find the value of $\Delta t$.
The temperature difference values are given. We get,
$ \Rightarrow \Delta t = {35.1^ \circ }C - {34.9^ \circ }C$
After subtracting we get,
$ \Rightarrow \Delta t = {0.2^ \circ }C$
Now we have got the values of $Q$ and $\Delta t$. We can substitute in the equation. we get,
$ \Rightarrow 300J = C \times {0.2^ \circ }C$
We need to find the capacity value. We get,
$ \Rightarrow C = \dfrac{{300J}}{{{{0.2}^ \circ }C}}$
We can simplify the values with the help of the division. We get,
$ \Rightarrow C = 1500J{/^ \circ }C$.
We have found the value of capacity. The value of the capacity is $1500J{/^ \circ }C$
$\therefore C = 1500J{/^ \circ }C$

Hence option \[\left( D \right)\] is the correct answer.

Note: Students will often get confused with the heat capacity and the specific heat capacity. The heat capacity can be defined as the ratio of heat energy transferred to an object. Specific heat capacity is the amount of heat needed to increase the temperature of $1g$ of a substance to ${1^ \circ }C$.