. please ans. this with complete solutions . If u don't know then no need to give faltu answers
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. please ans. this with complete solutions . If u don't know then no need to give faltu answers
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Answer:
39. 1050 w
40. 1.9L
Explanation:
39. the rating can be found out by calculating the total power usage
We know that Q=mc(t2-t1)
where Q= power , m= mass , t1 = initial temperature and t2= final temprature
c= specific heat of water =4.2kj/kg
given m= 18/hr=18×(60×60)=0.005kg/sec
t1=10°c and t2 = 60° c
therefore, Q= 0.005×4.2(60-10)=1.05kw
=1050W
40. V1=1000C.C.V2=?
T1=27C=300K;T2=297C=570K
Since the pressure is constant
∴T1V1=T2V2
∴3001000=570V2
∴V2=3001000×570=1900c.c.
=1.9litres
Yo
39.
we see that rating of the geyser ie its power will be equal to heat generated by it per second cause power=heat/time
heat=Q=mCdelT
delT=T1-T2
=60-10
=50
C of water is 4.2Kj/kg (this should be remembered by you..ratta maar lo)
m=18kg/hr
=18/60*60
=0.005kg/sec
Now power=heat=0.005*4.2*1000*50 W
this will give u answer as option C ie 1050W
40.
in this problem we have to just apply Charles law which can be given as
T1V2=T2V1
on plugging in the values we get
V2*300=570*1. (do not forget to convert Celcius to Kelvin)
V2=1.9L
hence the correct answer is option B
Hope this helps