Question

The resistance of an electric toaster has a temperature dependence given by $R(T)={{R}_{0}}\left[ 1+\alpha \left( T-{{T}_{0}} \right) \right]$ in its range of op eration. At ${{T}_{0}}=300K$, ${{R}_{0}}=100\Omega$ and at T = 500K, R= 120$\Omega$. The toaster is connected to a voltage source at 200V and its temperature is raised at a constant rate from 300K to 500K in 30s. The total work done in raising the temperature is:
$\text{A}\text{. }400\ln \dfrac{5}{6}J$
$\text{B}\text{. 2}00\ln \dfrac{2}{3}J$
$\text{C}\text{. 3}00J$
$\text{D}\text{. }400\ln \dfrac{1.5}{1.3}J$

Answer 1

Find the heat consumed by the resistance by using the formulas $P=\dfrac{{{V}^{2}}}{R}$ and $dE=P.dt$. Find the exact expression for the resistance with help of given data. Find the relation between time and temperature and then integrate $dE=P.dt$ for the time interval of 30 seconds.

Formula used:
$P=\dfrac{{{V}^{2}}}{R}$
$dE=P.dt$

Complete step by step solution:
The work done in raising the temperature will be negative of heat generated by the resistance of the toaster.
The power of energy consumed by a resistance is given as P = Vi.
Here P is the power, V is the potential difference across the resistance and i is the current flowing in the circuit.
Power is the energy consumed in one unit of time. Suppose the resistance consumes dE amount of energy in dt time. Then the power P will be,
$P=\dfrac{dE}{dt}$.
$\Rightarrow dE=P.dt$ …. (i).
Therefore, in time interval dt, the energy consumed by the resistance is equal to Pdt.
And for a circuit $P=\dfrac{{{V}^{2}}}{R}$
Hence, $dE=\dfrac{{{V}^{2}}}{R}dt$ …. (i).
Therefore, if we integrate equation (i) over the given interval of time then we will find the energy consumed in 30 seconds.
However, here the resistance of the circuit is a function of temperature and temperature of time.
Let us first find a complete expression for the resistance.
It is given that $R(T)={{R}_{0}}\left[ 1+\alpha \left( T-{{T}_{0}} \right) \right]$ ….. (ii).
And it is also given that ${{T}_{0}}=300K$ and ${{R}_{0}}=100\Omega$.
It is said that at temperature T = 500K the resistance $R=120\Omega$.
Substitute the ${{T}_{0}}=300K$, ${{R}_{0}}=100\Omega$, T = 500K and $R=120\Omega$.
$\Rightarrow 120=100\left[ 1+\alpha \left( 500-300 \right) \right]$
$\Rightarrow \dfrac{120}{100}=\left[ 1+\alpha \left( 200 \right) \right]$
$\Rightarrow \dfrac{6}{5}=1+200\alpha$
$\Rightarrow 200\alpha =\dfrac{6}{5}-1=\dfrac{1}{5}$
$\Rightarrow \alpha =\dfrac{1}{5\times 200}=\dfrac{1}{1000}={{10}^{-3}}{{K}^{-1}}$.
Therefore, the expression of the resistance can be written as $R(T)=100\left[ 1+{{10}^{-3}}\left( T-300 \right) \right]$.
……(iii)
Hence, $dE=\dfrac{{{V}^{2}}}{R}dt=dE=\dfrac{{{V}^{2}}}{100\left[ 1+{{10}^{-3}}\left( T-300 \right) \right]}dt$ …..(iv).
Now we will find the equation between temperature T and time t.
It is given that the temperature changes unformly from 300K to 500K in 30 seconds. Therefore, the graph of temperature versus time will be a straight line. An equation of a straight line is given as y = mx +c.
Here x=t and y=T.
Therefore, T = mt +c.
Let us calculate the value of m and c.
According to the given data. At time t=0, temperature T = 300K.
Therefore, $300=m(0)+c$
$\Rightarrow c=300k$.
At time t=30secs, T=500K.
Therefore, $500=m(30)+300$
$\Rightarrow 200=30m$
$\Rightarrow m=\dfrac{200}{30}=\dfrac{20}{3}K{{s}^{-1}}$.
Then the relation between temperature and time is $T=\dfrac{20}{3}t+300$.
Therefore, $T-300=\dfrac{20}{3}t$ …..(v).
Substitute the value of (T-300) in equation (iv)
$\Rightarrow dE=\dfrac{{{V}^{2}}}{100\left[ 1+{{10}^{-3}}.\dfrac{20}{3}t \right]}dt$
Integrate both the sides for a time interval of 30seconds.
$\int{dE}=\int\limits_{0}^{30}{\dfrac{{{V}^{2}}}{100\left[ 1+{{10}^{-3}}.\dfrac{20}{3}t \right]}dt}$
Here V=200V.
$\Rightarrow \int{dE}=\int\limits_{0}^{30}{\dfrac{{{200}^{2}}}{100\left[ 1+\dfrac{0.02}{3}t \right]}dt}$

$\Rightarrow \int{dE}=\dfrac{{{200}^{2}}}{100}\int\limits_{0}^{30}{\dfrac{dt}{\left[ 1+\dfrac{0.02}{3}t \right]}}$
$\Rightarrow E=\dfrac{{{200}^{2}}}{100}\ln \left. \left[ 1+\dfrac{0.02}{3}t \right] \right|_{0}^{30}$
$\Rightarrow E=400\ln \left[ 1+\dfrac{0.02}{3}(30) \right]$
$\Rightarrow E=400\ln \left[ 1+0.2 \right]$
$\Rightarrow E=400\ln \left[ 1.2 \right]=400\ln \left[ \dfrac{6}{5} \right]$
Therefore, work done in raising the temperature is $-400\ln \left[ \dfrac{6}{5} \right]=400\ln \left[ \dfrac{5}{6} \right]$.

Hence, the correct option is A.

Note: We can also integrate the energy with respect to temperature. Then the limits will be from 300K to 500K.
The relation between time and temperature can also be found in this way:
It is given that the temperature increases uniformly with time. Therefore, the graph of temperature versus time will be a straight line. In a time interval of 30 seconds the temperature increases by 200K, which that the slope of the graph will be $m=\dfrac{dT}{dt}=\dfrac{200K}{30s}=\dfrac{20}{3}K{{s}^{-1}}$.
$\Rightarrow dt=\dfrac{3}{20}dT$.
With this, we can then integrate with respect to temperature from 300K to 500K.