Question

0.5 kg of ice at 0°C is heated uniformly by an electric heater of power 2kW if all heat is absorbed by ice calculate the time intervals in seconds for (i) ice to completely melt to form water at 10°C (ii) water to attain a temperature of 100°C (iii) water to change to steam at 100°C given: Sp.latent heat of ice = 336000JKg^-1 Sp.latent heat of steam = 2260000JKg^-1 Sp.heat capacity of water = 4200JKg^-1K^-1​

Answer 1

Answer:

calculate the volume of O2 required for combustion of 11.2 litre butane the percentage purity of o2 is 60 percentage?

Charge on 1 electron= 1.6x10^-19 C

So, in 1 coulomb of charge 1/(1.6x10^-19)

= 6.25x10^18 ...

Answer 2

Answer:

m=0.5kg

P=2kW=2000W

L(ice)=336000

L(steam)=2260000

c(water)=4200

(i)Power given by electric heater=Energy required to convert ice completely into water at 10°C

T(final)=10°C

T(initial)=0°C

Time taken=t

P×t=mL(ice) + mc(T(final)-T(initial))

2000×t=0.5×336000+0.5×4200×(10-0)=189000

t=94.5s

(ii)Time taken =t

T(final)=100°C

T(initial)=0°C

Power given by electric heater=Energy required to heat water to 100°C

P×t=mc(T(final)-T(initial))

2000×t=0.5×4200×(100-0)=210000

t=105s

(iii)Time taken=t

L(steam)=2260000

Power given by electric heater=Energy required to convert water at 100°C to steam at 100°C

P×t=mL(steam)

2000×t=0.5×2260000=1130000

t=565s