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Step-by-Step Solution
Step 1: Identify the temperature coefficient of resistance
The resistance of the toaster varies with temperature as
$R(T) = R_0 \bigl[1 + \alpha \,(T - T_0)\bigr].$
We are given:
$R(T_0=300\,\text{K})=100\,\Omega.$
$R(T=500\,\text{K})=120\,\Omega.$
Substitute these values into the temperature-dependence equation to find $\alpha$:
$120 = 100 \Bigl[1 + \alpha (500 - 300)\Bigr].$
That is,
$120 = 100\bigl[1 + 200\,\alpha \bigr].$
Divide both sides by 100:
$1.2 = 1 + 200\,\alpha \quad\Rightarrow\quad 200\,\alpha = 0.2 \quad\Rightarrow\quad \alpha = 10^{-3}\,\text{K}^{-1}.$
Step 2: Determine how temperature changes with time
The temperature of the toaster is increased uniformly from $300\,\text{K}$ to $500\,\text{K}$ in $30\,\text{s}$. Thus, the total temperature rise is $500 - 300 = 200\,\text{K}$ over $30\,\text{s}$. The rate of change of temperature is:
$\Delta T \;=\; \frac{200}{30} \;=\;\frac{20}{3}\;\text{K/s}.$
If $t$ is the time (in seconds), then the instantaneous temperature $T(t)$ after $t$ seconds is
$T(t) = 300 + \frac{20}{3}\,t.$
Hence, the change in temperature above $300\,\text{K}$ at time $t$ is
$T(t)\;-\;300 \;=\;\frac{20}{3}\,t.$
Step 3: Express the instantaneous resistance as a function of time
Using $R(T) = R_0\bigl[1 + \alpha\,(T-300)\bigr]$ and substituting $T(t)-300 = \tfrac{20}{3}\,t$, we get:
$R(t)\;=\;100 \Bigl[1 + 10^{-3} \Bigl(\frac{20}{3}\,t\Bigr)\Bigr]
\;=\;100\Bigl[1 + \frac{t}{150}\Bigr].$
Step 4: Write down the expression for total work done
The work done in time $dt$ when a constant voltage $V = 200\,\text{V}$ is applied across resistance $R(t)$ is given by the power $P = \tfrac{V^2}{R(t)}$ multiplied by $dt$. Hence, the total work done in $30\,\text{s}$ is:
$W = \int_{0}^{30} \frac{V^2}{R(t)}\,dt.$
Substitute $V = 200\,\text{V}$ and $R(t) = 100\bigl[1 + \tfrac{t}{150}\bigr]$:
$W
= \int_{0}^{30} \frac{(200)^2}{100\bigl(1 + \tfrac{t}{150}\bigr)}\,dt
= \int_{0}^{30} \frac{40000}{100\Bigl(1 + \tfrac{t}{150}\Bigr)}\,dt
= 400 \int_{0}^{30} \frac{dt}{1 + \tfrac{t}{150}}.$
Step 5: Evaluate the integral
Let $u = 1 + \frac{t}{150}.$ Then $du = \frac{1}{150}\,dt$ or $dt = 150\,du.$
When $t=0$, $u=1$; and when $t=30$, $u=1 + \tfrac{30}{150} = 1.2.$
Thus, the work integral becomes:
$W = 400 \int_{u=1}^{1.2} \frac{150\,du}{u}
\;=\; 400 \times 150 \int_{1}^{1.2} \frac{du}{u}
\;=\; 60000 \Bigl[\ln(u)\Bigr]_{1}^{1.2}
\;=\; 60000\,\ln\bigl(1.2\bigr).$
Since $1.2 = \tfrac{6}{5}$, we can write:
$W = 60000\,\ln\bigl(\tfrac{6}{5}\bigr)\;\text{J}.$
Final Answer
$\displaystyle \boxed{60000 \,\ln\bigl(\tfrac{6}{5}\bigr)\,\text{J}}$
