Question

What is the maximum amount of work that may be obtained by operating a heat engine between two beakers of water which are initially at 0 degree C and 100 degree C and which both contain 1 times 10^-3 m^3 of water?

Answer 1

Using a formula, we have

$\Delta$ Q = m_w C_w $\Delta$ T

$\Delta$ Q = ( _{$\rho$ w} V) C_w (T_f - T₀)

where, V = volume of water = 1 x 10^-3 m³

_{$\rho$ w} = density of water = 1000 kg/m³

C_w = specific heat constant of water = 4186 J/kg.⁰C

then, we get

$\Delta$ Q = [(1000 kg/m³) (1 x 10^-3 m³)] (4186 J/kg.⁰C) [(100 - 0) ⁰C]

$\Delta$ Q = 4.18 x 10⁵ J

The maximum amount of work that may be obtained by an operating heat enegine which given as :

we know that, $\Delta$ U = $\Delta$ Q - $\Delta$ W

$\Delta$ W = $\Delta$ Q                                                      (where, $\Delta$ U = internal energy = 0)

$\Delta$ W = 4.18 x 10⁵ J