Question

Liquid ammonia is used in ice factory for making ice from water. If water at 20 °C is to be converted into 2 kg ice at 0 °C, how many gram of ammonia are to be evaporated.
(Given :The latent heat of vaporisation of ammonia =341 cal /g) ​

Answer 1

❏ USED FORMULAS:-

Let, An object of mass m and which is at initial temperature $\bf t_1$ ,

Now if the temperature of the object is changed to a temperature $\bf t_2$ ,

(where the phase of the object is unchanged)

Then the Required Heat,

$\sf\longrightarrow \boxed{HEAT_s=m\times S\Delta t}$

where,

$\sf\rightarrow\Delta t=change\: in \:Temperature=(t_2-t_1)$

$\sf\rightarrow$ m=mass of the object.

$\sf\rightarrow$ S=specific Heat.

Let, An object of mass m and which is at initial temperature $\bf t_1$ ,

Now if the phase of the object is changed without change in temperature.

Then the Required Heat,

$\sf\longrightarrow \boxed{HEAT_f=m\times L_f}$

where,

where, $\sf\rightarrow$ m=mass of the object.

$\sf\rightarrow L_f$=specific Heat.

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❏ Solution:-

• Given➾ •mass(m)=2 kg=2000 gm

•initial temperature $\bf t_i$ =20° c

•final temperature $\bf t_f$ =O°C

•Latent heat of vap. of $NH_3$=341 cal/gm

•To Find➾ how many gram of ammonia are to be evaporated?

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➾ Amount of heat energy released in cooling 2 kg =water from 20°C to 0°C.

=(2000×1×20)cal

[∵ specific heat of water=1 cal/gm-°C]

=40,000 cal

➾ Amount of heat energy released in converting 2kg =2000gm of water at 0°C. to ice.

=(2000×80) cal

[∵Latent Heat of ice =80cal/gm]

=1,60,000 cal

➾Therefore total energy required in converting 2kg of water from 20°C to ice.

=(40,000+1,60,000) cal

=2,00,000 cal

➾ Therefore, grams of Ammonia to be evaporated is

=$\Large{\frac{2,00,000}{341}\:gm}$

= $\boxed{\sf\red{586.5 gm}}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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