A metal ball of mass 1kg is heated by means of a 20W heater in a room at `20^(@)C` . The temperature of the ball becomes steady at `50^(@)C` . (a) Find the area of loss of heat to the surrounding when the ball is at `-50^(@)C` . (b) Assuming Newton's law of cooling, calculate the rate of loss of heat to the surrounding when the ball is at `30^(@)C` . (c) Assume that the temperature of the ball rises uniformly from `20^(@)C` to `30^(@)C` in 5 minutes. Find the total loss of heat to the surrounding during this period. (d) Calculate the speific heat capicity of the metal.

Answer

A, B, C, D

Explanation

Let the water equivalent of calorimeter
`=m`
In 1st case,
`((m+50 xx 10^(-3) xx 4200 xx 5)/(10)`
`=rate of heat flow`
In second case,
`((m + 100 xx 10^(-3) xx 4200 xx 5)/(18)`
`= rate of heat flow`
From (i) and (ii),
`((m+50 xx 10^(-3) xx 4200 xx 5)/(10)`
`= ((m+100 xx 10^(-3) xx 4200 xx 5)/(18)`
`implies (m + 50 xx 10^(-3)) xx 18`
`= 10m + 1000 xx 10^(-3)`
`implies 18 m + 18 xx 50 xx 10^(-3)`
`= 10m + 1000 xx 10^(-3)`
`implies 8m = 100 xx 10^(-3) kg`
`m = 12.5 xx 10^(-3) kg = 12.5 gms`.