A metal drill of power output 500W drills a hole in a lead cube of mass 0.25 kg in 6.5 sec. The
specific heat capacity of lead is 130J kg^-1 K^-1 Calculate:
(i) The heat generated by the metal drill in 1 sec.
(ii) The heat generated by the metal drill in 6.5 sec.
(i) If t°C is the rise in temperature of the lead cube, heat absorbed by the cube in terms of t.
(iv) The value of t. Assume that all the heat generated by the drill is absorbed by the cube.
please someone answer ..
very urgent it would be really helpful if anyone here can solve
Answers
Heat generated by the metal drill in 1 sec = 15.38°C
Heat generated in 6.5 sec = 99.97°C
Explanation:
Given: Specific heat capacity of lead is 130J/kgK
Power output = 500 W
lead cube of mass = 0.25 kg
Time = 6.5 second.
Find:
(i) The heat generated by the metal drill in 1 sec.
(ii) The heat generated by the metal drill in 6.5 sec.
Solution:
We know that Conversion of Electrical Energy into Thermal Energy is denoted by the formula:
P * t = mcθ
where P is power and t is time in seconds, m is mass in kg, c is specific heat capacity, and θ is temperature change in Celsius.
(i) 500 * 1 = 0.25* 130 * θ
Therefore θ = 500 / (0.25 *130)
θ = 15.38°C
Heat generated by the metal drill in 1 sec = 15.38°C
(ii) Heat generated by the metal drill in 6.5 sec = 15.38 * 6.5 = 99.97°C
Explanation:
(1) power =work done/time
workdone =heat,H=p*t=500*6.5=3250 joule for6.5 seconds
heat per second=3250/6.5 =500 joule
(2) H=3250joule
(3) heat lost by metal drill = heat absorbed by lead cube
heat lost by metal drill= mass of lead *specific heat of lead*temperature rise
heat lost by metal drill =0.25*130*t=32.5t joule
(4) change in temperature=?
we know that
Heat lost by metal drill= heat gained by lead cube
p*t= 0.25*130*change in temperature
500*6.5=32.5*change in temperature
change in temperature=(500*6.5)/(32.5)=100 degree Celsius
change in temperature= 100 degree celsius