Respuesta :
Answer: The mass of ice you would need to add to bring the equilibrium temperature of the system to 300 K is [tex]16.14 \times 10^{4}[/tex] kg.
Explanation:
We know that relation between heat energy and specific heat is as follows.
q = [tex]m \times S \times \Delta T[/tex]
As density of water is 1 kg/L and volume is given as 400,000 L. Therefore, mass of water is as follows.
Mass of water = Volume × Density
= [tex]400,000 L \times 1 kg/L[/tex]
= 400,000 kg
or, = [tex]400,000 \times 10^{3}[/tex] g (as 1 kg = 1000 g)
Specific heat of water is 4.2 J/gm K. Therefore, change in temperature is as follows.
[tex]\Delta T[/tex] = 305 K - 273 K
= 32 K
Now, putting the given values into the above formula and calculate the heat energy as follows.
q = [tex]m \times S \times \Delta T[/tex]
= [tex]400,000 \times 10^{3} \times 4.2 \times 32 K[/tex]
= [tex]5376 \times 10^{7}[/tex] J
or, = [tex]5376 \times 10^{4}[/tex] kJ
According to the enthalpy of melting of ice 333 kJ/Kg of energy absorbed by by 1 kg of ice. Hence, mass required to absorb energy of [tex]5376 \times 10^{4}[/tex] kJ is calculated as follows.
Mass = [tex]\frac{5376 \times 10^{4} kJ}{333 kJ/Kg} \times 1 kg[/tex]
= [tex]16.14 \times 10^{4}[/tex] kg
Thus, we can conclude that the mass of ice you would need to add to bring the equilibrium temperature of the system to 300 K is [tex]16.14 \times 10^{4}[/tex] kg.
