The heat capacity of liquid water is \[75.6\,J/K -1 mol\] , while the enthalpy of fusion of ice is \[6.0\,Kj/mol\]. What is the smallest number of ice cubes at \[{0^o}C\], each containing \[9.0{\text{ }}g\] of water, needed to cool \[500{\text{ }}g\] of liquid water from \[{20^o}C\] to ${0^0}C$?
(A) $1$
(B) $7$
(C) $14$
(D) $21$
Answer
595.5k+ views
Hint: The Heat capacity of a matter is known as the amount of heat needed to bring a unit change in the temperature of that matter. Its SI unit is Joule per kelvin. In the given question we need to apply the first law of thermodynamics to calculate the desired number of ice cubes that are needed to cool \[500{\text{ }}g\] of liquid water from \[{20^o}C\] to ${0^0}C$. For that, we will need to calculate the total heat lost by water and total heat gain by ice.
Complete step by step answer:
Given,Heat capacity of water \[ = 75.6\,J/K -1 mol\]
Enthalpy of fusion of ice \[ = 6.0\,Kj/mol\]
Mass of each ice cube \[ = 9{\text{ }}g\]
Mass of liquid water \[ = 500{\text{ }}g\]
Heat Capacity is given as:
$Q = mc\Delta T$
Where \[m = \]Mass , \[c = \]Heat capacity of the matter and \[\Delta T = \]Change in Temperature
No. of Moles of Liquid water is given as,
$No.\,\,of\,moles = \dfrac{{given\,mass}}{{molar\,mass}} = \dfrac{{500}}{{18}} = 27.78$
Heat Capacity of water is given as,
\[Q\;(heat{\text{ }}lost)\; = mc\Delta T\]
\[Q = 27.78 \times 75.6 \times (293 - 273) = 42000\,J\]
$Q = \dfrac{{42000}}{{1000}} = 42\,Kj$
No. of Moles of ice is given as,
$No.\,\,of\,moles(1\,ice\,cube) = \dfrac{{given\,mass}}{{molar\,mass}} = \dfrac{9}{{18}} = 0.5$
We have Enthalpy of fusion of ice \[ = 6.0\,Kj/mol\] so now,
Enthalpy of fusion of $1$ ice cube$ = 0.5 \times 6 = 3\,Kj$
Let there are $'n'$ number of ice cubes so Enthalpy of fusion of $'n'$ ice cubes$ = 3n\,Kj$
Now, By First Law of Thermodynamics ;
$Heat\,lost = Heat\,gain\,$
\[42 = 3n\]
$n = 14$
Hence, The total No. of ice cubes required is $14$.
Therefore, Option (C) is correct.
Note:The enthalpy of fusion is defined as the amount of heat needed to transform the unit mass of a solid at its melting point into a liquid without a rise in temperature of the substance. The enthalpy of fusion is also called the latent heat of fusion. While putting the values in the formula always convert all the units into the standard form and correctly calculate the number of moles.
Complete step by step answer:
Given,Heat capacity of water \[ = 75.6\,J/K -1 mol\]
Enthalpy of fusion of ice \[ = 6.0\,Kj/mol\]
Mass of each ice cube \[ = 9{\text{ }}g\]
Mass of liquid water \[ = 500{\text{ }}g\]
Heat Capacity is given as:
$Q = mc\Delta T$
Where \[m = \]Mass , \[c = \]Heat capacity of the matter and \[\Delta T = \]Change in Temperature
No. of Moles of Liquid water is given as,
$No.\,\,of\,moles = \dfrac{{given\,mass}}{{molar\,mass}} = \dfrac{{500}}{{18}} = 27.78$
Heat Capacity of water is given as,
\[Q\;(heat{\text{ }}lost)\; = mc\Delta T\]
\[Q = 27.78 \times 75.6 \times (293 - 273) = 42000\,J\]
$Q = \dfrac{{42000}}{{1000}} = 42\,Kj$
No. of Moles of ice is given as,
$No.\,\,of\,moles(1\,ice\,cube) = \dfrac{{given\,mass}}{{molar\,mass}} = \dfrac{9}{{18}} = 0.5$
We have Enthalpy of fusion of ice \[ = 6.0\,Kj/mol\] so now,
Enthalpy of fusion of $1$ ice cube$ = 0.5 \times 6 = 3\,Kj$
Let there are $'n'$ number of ice cubes so Enthalpy of fusion of $'n'$ ice cubes$ = 3n\,Kj$
Now, By First Law of Thermodynamics ;
$Heat\,lost = Heat\,gain\,$
\[42 = 3n\]
$n = 14$
Hence, The total No. of ice cubes required is $14$.
Therefore, Option (C) is correct.
Note:The enthalpy of fusion is defined as the amount of heat needed to transform the unit mass of a solid at its melting point into a liquid without a rise in temperature of the substance. The enthalpy of fusion is also called the latent heat of fusion. While putting the values in the formula always convert all the units into the standard form and correctly calculate the number of moles.
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