Question

a black body at 327 degree celsius suspended enclosure at 27 degrees celsius cools at certain rate. If temperature of black body is lowered to t' degree celsius so that rate of cooling becomes half of initial value then find approximate value of t' in degree celsius

Answer 1

Answer:

t = 231.53 °C

Explanation:

As we know that rate of cooling is directly proportional to the forth power of temperature.

dT/dt ∝ T⁴        -------------1

Given that

T₁ = 327 C = 327 + 273  = 600 K

When rate of cooling become half then temperature is t

(dT/dt)/2 ∝ t⁴            ----------2

From 1 and 2

T⁴  = 2 t⁴

600⁴ = 2 .t⁴

t=504.53 K

t = 504.3 - 273 =231.53 °C

t = 231.53 °C