Question

An object at 27°C has its temperature increased to 37°C.

The power then radiated by this object increases by how many percent?
a. 3.3
b. 14
c. 37
d. 253

Answer 1

b. 14

Step-by-step explanation:

`T_(i)` = Initial temperature = 27 °C = 27 + 273 = 300 K

`T_(f)` = Final temperature = 37 °C = 37 + 273 = 310 K

`P_(i)` = Initial Power radiated by the object

`P_(f)` = Final Power radiated by the object

We know that the power radiated is directly proportional to fourth power of the temperature. hence

`(P_(f))/(P_(i)) = (T_(f)^(4) )/(T_(i)^(4) )\\(P_(f))/(P_(i)) = ((310)^(4) )/((300)^(4) )\\(P_(f))/(P_(i)) = 1.14\\P_(f) = (1.14) P_(i)`

Percentage increase in power is given as

`((P_(f) - P_(i))*100)/(P_(i)) \\(((1.14) P_(i) - P_(i))*100)/(P_(i)) \\14`