A metal block is placed in a room which is at `10^(@)C` for long time. Now it is heated by an electric heater of power `500W` till its temperature becomes `50^(@)C` Its initial rate of rise of temperature is `2.5^(@)C//sec` The heater is switched off and now a heater of `100W` is required to maintain the temperature of the block at `50^(@)C` The heat radiated per second when the block was `30^(@)C` is given as `alpha` watt. Find the value of `((alpha)/(10))` (Assume Newton's law of cooling to be valid .

Answer

5

Explanation

Given at `50^(@)C` Newton's law of cooling Power loss
`100 = k (50 -10) rArr k = (10)/(4) = (5)/(2)`
`:.` At `30^(@)C P_(LOSS) =k (30 -10)` .