A very large mass M of hot porous rock equal to 10^12 kg is to be utilized to ge
ID: 983595 • Letter: A
Question
A very large mass M of hot porous rock equal to 10^12 kg is to be utilized to generate electricity by injecting water and using the resulting hot steam to drive a turbine. As a result of heat extraction, the temperature of the rock drops according to Q = -MC_p dT/dt, where C_p is the specific heat of the rock which is assumed to be independent of temperature. If the plant produces 1.36 times 10^9 kW hr of energy per year, and only 25 percent of the heat extracted from the rock can be converted to work, how long will it take for the temperature of the rock to drop from 600degreeC to 110degreeC? Assume that for the rock C_P = 1 J/(g K).Explanation / Answer
Kw hr= 36*105 joules
1.36*109 Kwhr= 1.36*109* 36*105 joules= 48.96*1014 joules
but only 25% of the heat is converted to work. So total heat produced/ year= (48.96/0.25)*1014 joules=195.84*1014 jouules per year
1 year= 365 days =365*24 hrs= 365*24*60*60 seconds =31536000 sec
heat produced/sec= 195.84*1014/ 31536000=621004566 joule/sec
This is equal to -mCpdT/dt= 621004566 joule/sec
which on integration givee
-mcp*delT= 621004566 t ,where t = time required
given m=mass of rock= 1012 kg -dT= 600-110= 490
Cp= 1 j/g.k= 1000 j/kg. K
1012*1000*(490)= 6210004566*t
t= 1012*1000*490/6210004566=7.89*108 seconds =25 years.