Question

10 kilograms of water are boiled at \( 373 K \), at a pressure of \( 1.013 \times 10^{5} N / m ^{2} \) and converted into steam. The specific latent heat of vaporization of water is \( 539 kcal / kg .1 L \) of water, on conversion into steam, occupies \( 1671 L \). The mechanical equivalent of heat is \( 4186 J / kcal \). Calculate 

(a) the energy supplied to the system (water)

(b) the work done by the system

(c) the change in the internal energy of the system.

Answer 1

Given P = 1.013 x 10⁵ N/ m²

M = 10 kg

L = 539 kcal/kg

V₁ = 1 L

V₂ = 1671 L

(a) Q = ML

= 10 x 10³ x 539

Q = 5390 kcal

(b) Work done

W = Pdv

= 1.013 x 10⁵ (V₂- V₁)

= 1.013 x 10⁵ (1671 -1) x 10^-3

W = \(\frac {1691.71 \times 10^2\,J}{4.2}\)

W = 402.7 x 10²

W = 40.27 kcal

(c) dθ = du + dW

du = dθ - dW

= 5390 - 40.27

du = 5349.73 kcal