Question

The following data is obtained during boys gas calorimeter experiment volume of gas used = 0.50 m3 at NTP weight of water heated = 125 kg temp of inlet water = 22.5 C temp of outlet water = 42.5 C weight of steam condensed = 0.129 kg. Calculate G.C.V. per m3 at STP provided that heat liberated in water vapor and cooling the condensate as 587 kcal/ m3.​

Answer 1

Answer:

sorry but i dont know

Answer 2

Answer:

Gross calorific values at STP for water and steam are $2.935 \times {10^6}Kcal/{m^3}$ and $3.02892 \times {10^3}Kcal/{m^3}$ respectively.

Explanation:

As per the question we have given

The volume of gas used = 0.5 $m^3$

The weight of water heated (m) = 125 kg

The temperature of inlet water ($T_1$) = 22.5°C

The temperature of outlet water($T_2$) = 42.5°C

Weight of steam condensed (m) = 0.129 kg

Specific heat of the water and steam (S) = 587Kcal/$m^3$

Gross calorific value for water

$\[\begin{gathered} Q = mS\Delta T \hfill \\ Q = 125 \times 587 \times \left( {42.5 - 22.5} \right) \hfill \\ Q = 1467500\ Kcal \hfill \\ \end{gathered} \]$

Gross calorific value per $m^3$ for water = $\[\frac{{1467500}}{{0.5}} = 2935000 = 2.935 \times {10^6}Kcal/{m^3}\]$

The gross calorific value of steam

$\[\begin{gathered} Q = mS\Delta T \hfill \\ Q = 0.129 \times 587 \times \left( {42.5 - 22.5} \right) \hfill \\ Q = 1514.46 Kcal \hfill \\ \end{gathered} \]$

Gross calorific value per $m^3$ for steam = $\[\frac{{1514.46}}{{0.5}} = 3028.92 = 3.02892 \times {10^3}Kcal/{m^3}\]$

Therefore, gross calorific values at STP for water and steam are $2.935 \times {10^6}Kcal/{m^3}$ and $3.02892 \times {10^3}Kcal/{m^3}$ respectively.