Fig 91 shows four plastic containers P Q R and S with differ
Solution
Ans(b)
Based on the information given, box S has highest specific heat capacity and low thermal conductivity. So though
it is smaller in size than box R, it will be economic and convenient to store the fish.
Ans (c)
(i) Thermal energy gained by ice and water= energy needed to melt 2.5Kg ice at 00C
+ energy needed to raise the temp of water from 00 C to 100c
= mL +mst
= m(L +st)
= 2.5 kg [ 334000 J/Kg + 4200 X 10 K/Kg]
=940000 J
(ii) Thermal enrgy lost by the fish = mass of fish(M) X its sp heat X its temperatre fall
= 15Kg X 3400 J/Kg/K X 15 K
= 76500 J
(iii) Thermal energy lost by the plastic container = mass of container X its sp heat X temp fall
= 5Kg X 1050 J/Kg/K X 15 K
= 78750 J
PLEASE NOTE this question paper contains two sets of questions in : total 1(in b)+ 5 (in c) components.
(iv) thermal energy that flows through the wall in t hours= heat transfer by wall in hour X time in hour
Now the plastic wall tranfers heat at = 120 J/ min (given) = 120 X 60 J/hr
The required energy flow throgh wall = 7200 t J in t hr.
(v) The time taken to reach equlibrium tempreature = total heat loss/ rate of loss
(vi) The answer (v) is higher than the actual value as the loss of enrgy from the container in the surroundings also affect.
