An experiment takes 20 min to raise the temperature of 1kg of water in a container from 20° C to 100° C and another 60 min to convert it totally into steam by a heater supplying heat at a uniform rate. Neglecting the specific heat of the container and taking the specific heat of water to be 1 cal/g-°C, the heat of vaporization will be:

CONCEPT:

• Specific heat: It is defined as the amount of heat required to increase the temperature by 1°K for 1 kg of material.

⇒ Q = C m ∆t     -----(1)

Where Q(J) = quantity of heat absorbed by a body, m(kg) = mass of the body, ∆t(°K) = Rise in temperature, and C = Specific heat

• S.I unit of specific heat is t is J kg-1 K-1.

CALCULATION:

Given m = 1kg, T₁ = 20°C, T2 = 100°C, t₁= 20 min = 1200 sec, t₂ = 60 min = 3600 sec and C = 1 cal/g-°C

C = 1 cal/g-°C = 4200 J/Kg-°C

• The heat supply by the heater to raise the temperature from 20°C to 100°C is given as,

⇒ Δt = 100 - 20 = 80°C

⇒ Q = CmΔt

⇒ Q = 4200×1×80

⇒ Q = 336×10³ J    -----(1)

• The rate of heat supplied by the heater is given as,

\(⇒ P=\frac{Q}{t}\)

\(⇒ P=\frac{336×10^3}{1200}\)

⇒ P = 280 J/sec     -----(2)

• The heat of vaporization is given as,

⇒ Q_v = P × t₂

⇒ Qv = 280 × 3600

⇒ Qv = 1008 × 10³J

• Hence, option 4 is correct.