If a body coated black at 600 K surrounded by atmosphere at 300K has cooling rate r, then find the rate of cooling of body at 900K and same surrounding temperature.
Answers
Answered by
16
Newton's law of cooling:
rate of cooling: dQ(t)/dt = h A [ T(t) - T_env ]
h = heat transfer coefficient
A = area of contact/ area for heat flow
Q(t) heat dissipated/radiated in Joules
Rate of cooling is proportional to temperature difference between the body and the surroundings.
rate of cooling at 900 °K = r * (900 - 300) /(600 - 300) = 2 r
rate of cooling is Twice.
rate of cooling: dQ(t)/dt = h A [ T(t) - T_env ]
h = heat transfer coefficient
A = area of contact/ area for heat flow
Q(t) heat dissipated/radiated in Joules
Rate of cooling is proportional to temperature difference between the body and the surroundings.
rate of cooling at 900 °K = r * (900 - 300) /(600 - 300) = 2 r
rate of cooling is Twice.
kvnmurty:
click on red heart thanks
Similar questions