Question

An aluminium container of mass 100 gm contains 200 gm of ice at - 20°C. Heat is added to the system at the rate of 100 cal/s. Find the temperature of the system after 4 minutes (specific heat of ice = 0.5 and L = 80 cal/gm, specific heat of A1 = 0.2 cal/gm/°C)Ans is 25.5 ° C

Answer 1

after 4 minutes heat added to system =100 x 4 x 60=24000 cal

first we check ice completely melt or not
H1=msdT=200 x 0.5 x 20
=2000cal
H2=msdT=100 x 0.2 x 20
=400 cal

hence ice is melted
heat use in melting of ice =mL
=200 x 80
=16000cal
now temperature increases due to rest heat =24000-(2400+16000)=5600cal
when ice is melted then they form water so we use now S=1cal/gm/c

now ,
heat loss =heat gain
5600=msT +MST
=(100 x 0.2 +200 x 1 )T
=(20+200) T
T=5600/220=560/22=25.5 C (approx)

Answer 2

The energy provided to the system in 4 mins will be utilized in 3 ways:
1. to increase the temperature of the system form -20 C to 0 C.
2. once it reaches to 0 C, the energy will be used to melt all the ice
3. to again increase the temp

for step 1:
Q=(mass of ice * specific heat of ice * change in temp)+(mass of Al * specific heat of Al * change in temp)