The resistance of an electrical toasterf has a temeprature dependence given by `R(T)=R_0[(T-T_0)]` in its range of operation. At `T_0=300 K,R=100Omega and at T=500 K,R=120 Omega`. The toasterf is connected to a voltage source at 200 V and its temperature is raised at a con- stant rate from 300 to 500 K in 30 s. The total work done in raising the temperature is :

Options

A. `400 ln ((1.5)/(1.3)) J`
B. `200 ln ((2)/(3)) J`
C. `60,000 ln ((6)/(5)) J`
D. 300 J

Answer

C

Explanation

`R= R_(0) [1 + alpha (T-T_(0))]`
Substituting given values in above relation we can write the following:
`120 = 100 [1 + alpha(500- 300)]`
Solving we get `alpha = (1)/(1000)` per K
Temperature increases from 300K to 500K in 30 seconds, linearly. We can write the temperature as a function of time as follows:
`T = T_(0) + (500-300)/(30) t`
Note that in above relation `T_(0)` is 300 `rArr T- T_(0) = (20t)/(3)`
Substituting in relation of resistance we get the following
`R= 100 [1 + (1)/(1000) xx (20t)/(3)]`
`rArr R = 100 + (2t)/(3)`
Now we can find work done by the battery during this interval of 30s
`W_(b) = underset(0)overset(30)int (E^(2))/(R ) dt= underset(0)overset(30)int ((200)^(2))/(100 + (2t)/(3)) dt = 120000 underset(0)overset(30s)int (dt)/(300+ 2t)`
`rArr W_(b) = 120000 underset(0)overset(30s)int (dt)/(300+ 2t)`
`rArr W_(b) = 120000 [(1)/(2) ln (300+ 2t)]_(0)^(30)`
`rArr W_(b) = 60000 ln. (6)/(5)J`
Hence option (c) is correct