In an industrial process $10 k g$ of water per hour is to be heated from $20^{\circ} C t o 80^{\circ} C .$ To do this steam at $150^{\circ} C$ is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at $90^{\circ} C .$ then___ $k g$ of steam is required per hour.
(Specific heat of steam = 1 calorie per gm°C, Latent heat of vaporization = 540 cal/gm)

Open in App

Solution

Heat required by 10 kg water to change its temperature from $20^{\circ} C t o 80^{\circ} C$ in one hour is $Q_{1} = (m c Δ T)_{w a t e r} = (10 \times 10^{3}) \times 1 \times (80 - 20) = 600 \times 10^{3} c a l o r i e$
In condensation (i) Steam release heat when it looses it's temperature from $150^{\circ} C t o 100^{\circ} C .$
$[m c_{s t e a m Δ T}]$
(ii) At $100^{\circ} C$ it converts into water and gives the latent heat. [mL]
(iii) Water release heat when it looses it's temperature from $100^{\circ} C t o 90^{\circ} C .$
$[m c_{s t e a m Δ T}]$
If m gm steam condensed per hour, then heat released by steam in converting water of $90^{\circ} C$
$Q_{2} = (m c Δ T)_{s t e a m} + m L_{s t e a m} + (m s Δ T)_{w a t e r} = m [1 \times (150 - 100) + 540 + 1 \times (100 - 90)] = 600 m c a l o r i e$
According to problem $Q_{1} = Q_{2} = m = 10^{3} g m = 1 k g = 1000 g r a m s .$

Suggest Corrections

Similar questions

Q. The amount of heat required to convert 1 g of ice (specific 0.5 cal at

g^{- 1 o} C^{- 1}

) at

- 10^{0} C

to steam at

100

^{\circ} C

is ___________.

[ Given: Latent heat of ice is

80 C a l / g m,

Latent heat of steam is

540 C a l / g m

, Specific heat of water is

1 C a l / g m / C

]

Steam at 100⁰ C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15⁰ C till the temperature of the calorimeter and its contents rise to 80⁰ C. The mass of the steam condensed in kilogram is (latent heat of vaporization of water is 540 cal/gm)

Ratio of heat energy required to change in states of $100 g$ of water at $100^{\circ} C$ to steam to $540 g$ of ice to water respectively is $5 : y$ then $y$ = ___

Take latent heat of fusion $80 c a l / g m$ and latent heat of vaporization as $540 c a l / g m$

Determine the final result when 200 gm of water and 20 gm of ice at $0^{\circ} C$ are in a calorimeter having a water equivalent of 30 gm. 50 gm of steam is passed into it at $100^{\circ} C$ . Take latent heat of fusion 80 cal/gm and latent heat of vaporization as 540 cal/gm

Steam at $100^{\circ} C$ is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at $15^{o} C$ till the temperature of the calorimeter and its contents rises to $80^{\circ} C .$

What will be the mass of the steam condensed (in kg)?

Take:

Specific heat of water $= 1 c a l g m^{- 1}$

Latent heat of vaporization $= 540 c a l g m^{- 1}$

Join BYJU'S Learning Program