Answer
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Hint: -Calculate the amount of heat energy released in cooling the ice from $20^\circ C$ to $0^\circ C$ and then the amount of heat energy released in converting $2kg$ water into ice at $0^\circ C$. Add them to get the total heat energy released in the process. Now calculate the amount of ammonia evaporated by the equality, Heat loss = Heat gain.
Formula used:
The energy released in decreasing the temperature = ${m_1}c\Delta t$ where ${m_1}$ = mass of a substance, $c$ = specific heat capacity of substance, and $\Delta t$ = change in temperature.
The energy released in converting water to ice = ${m_1}L$ where ${m_1}$ = mass of substance, and $L$ = latent heat of fusion of ice.
Step by step solution:
We know that the mass of water will not change while the shift of state. Let it be ${m_1}$. Therefore, ${m_1}$ = $2kg$
Now, Specific heat capacity of water, $(c)$ = $4200J/kg^\circ C$
Change in temperature, $(\Delta t)$ = $(20 - 0)$ = $20$
Therefore, the energy released in decreasing the temperature of water from $20^\circ C$ to $0^\circ C$ = ${m_1}c\Delta t$
${E_1}$= $2 \times 4200 \times 20$ = $168000J$
We know that the Latent heat of fusion of ice, $(L)$ = $336000J/kg$
Energy released in converting $2kg$ water to ice at $0^\circ C$ = ${m_1}L$
${E_2}$= $2 \times 336000$ = $672000J$
Now, total energy released in the reaction $(E)={E_1}+{E_2}$
$E$= $168000J$ + $672000J$ = $840000J$
Latent heat of evaporation of ammonia = $341cal/g$ = $341 \times 4184J/kg$ = $1426744J/kg$ % $(\because 1cal/g = 4184J/kg)$
Therefore, the energy released by $mkg$ of ammonia = $m$ × $1426744J/kg$
Now, we know that heat energy given out by ammonia will be equal energy taken by water that is the total energy released in the reaction $E$ ,
⇒ $1426744 \times m = 840000$
⇒ $m$ = $\dfrac{{840000}}{{1426744}}$
Which gives, $m = 0.58875kg$ = $588.75g$
Note:Calculate heat released in decreasing the temperature and heat released in converting to ice separately as different formulae are used for both the cases. Convert value to SI units before using them. Don’t forget to add units in the final answer.
Formula used:
The energy released in decreasing the temperature = ${m_1}c\Delta t$ where ${m_1}$ = mass of a substance, $c$ = specific heat capacity of substance, and $\Delta t$ = change in temperature.
The energy released in converting water to ice = ${m_1}L$ where ${m_1}$ = mass of substance, and $L$ = latent heat of fusion of ice.
Step by step solution:
We know that the mass of water will not change while the shift of state. Let it be ${m_1}$. Therefore, ${m_1}$ = $2kg$
Now, Specific heat capacity of water, $(c)$ = $4200J/kg^\circ C$
Change in temperature, $(\Delta t)$ = $(20 - 0)$ = $20$
Therefore, the energy released in decreasing the temperature of water from $20^\circ C$ to $0^\circ C$ = ${m_1}c\Delta t$
${E_1}$= $2 \times 4200 \times 20$ = $168000J$
We know that the Latent heat of fusion of ice, $(L)$ = $336000J/kg$
Energy released in converting $2kg$ water to ice at $0^\circ C$ = ${m_1}L$
${E_2}$= $2 \times 336000$ = $672000J$
Now, total energy released in the reaction $(E)={E_1}+{E_2}$
$E$= $168000J$ + $672000J$ = $840000J$
Latent heat of evaporation of ammonia = $341cal/g$ = $341 \times 4184J/kg$ = $1426744J/kg$ % $(\because 1cal/g = 4184J/kg)$
Therefore, the energy released by $mkg$ of ammonia = $m$ × $1426744J/kg$
Now, we know that heat energy given out by ammonia will be equal energy taken by water that is the total energy released in the reaction $E$ ,
⇒ $1426744 \times m = 840000$
⇒ $m$ = $\dfrac{{840000}}{{1426744}}$
Which gives, $m = 0.58875kg$ = $588.75g$
Note:Calculate heat released in decreasing the temperature and heat released in converting to ice separately as different formulae are used for both the cases. Convert value to SI units before using them. Don’t forget to add units in the final answer.
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