A 0.2-L glass of water at 20°C is to be cooled with ice to 5°C. Determine how much ice needs to be added to the water, in grams, if the ice is at 0°C. Also, determine how much water would be needed if the cooling is to be done with cold water at 0°C. The melting temperature and the heat of fusion

Question

A 0.2-L glass of water at 20°C is to be cooled with ice to 5°C. Determine how much ice needs to be added to the water, in grams, if the ice is at 0°C. Also, determine how much water would be needed if the cooling is to be done with cold water at 0°C. The melting temperature and the heat of fusion of ice at atmospheric pressure are 0°C and 333.7 kJ/kg, respectively, and the density of water is 1 kg/L.

Accepted Answer

A 0.2-L glass of water at 20°C is to be cooled with ice to 5°C. The amounts of ice or cold water that needs to be added to the water are to be determined.

Assumptions 1 Thermal properties of the ice and water are constant. 2 Heat transfer to the water is negligible.

Properties The density of water is 1 kg/L, and the specific heat of water at room temperature is [latex] C= 4.18 kJ / kg \cdot{ }^{\circ} C [/latex] (Table A-9). The heat of fusion of ice at atmospheric pressure is 333.7 kJ/kg,

Analysis The mass of the water is

[latex] m_{w}= ho V=(1 kg / L )(0.2 L ) = 0.2 kg [/latex]

We take the ice and the water as our system, and disregard any heat and mass transfer. This is a reasonable assumption since the time period of the process is very short. Then the energy balance can be written as

[latex] \underbrace{E_{i n}-E_{ ext {out }}}_{\begin{array}{c} ext { Net energy transfer } \ ext { by heat, work, and mass }\end{array}}=\underbrace{\Delta E_{ ext {system }}}_{\begin{array}{c} ext { Change in internal, kinetic, } \ ext { potential, etc.energies }\end{array}} [/latex]

[latex] ightarrow 0=\Delta U ightarrow(\Delta U)_{ ext {ice }}+(\Delta U)_{ ext {water }}=0 [/latex]

[latex] \left\lfloor m C\left(0^{\circ} C -T_{1} ight)_{ ext {solid }}+m h_{i f}+\left.m C\left(T_{2}-0^{\circ} C ight)_{ ext {liquid }} ight|_{ ext {ice }}+\left[m C\left(T_{2}-T_{1} ight) ight]_{ ext {water }}=0 ight. [/latex]

Noting that [latex] T_{1, ext { ice }}=0^{\circ} C ext { and } T_{2}=5^{\circ} C [/latex] and substituting

[latex]m\left[0+333.7 kJ / kg +\left(4.18 kJ / kg \cdot{ }^{\circ} C ight)(5-0){ }^{\circ} C ight]+(0.2 kg )\left(4.18 kJ / kg \cdot{ }^{\circ} C ight) [/latex]

[latex] (5 - 20)[/latex]

It gives [latex] m=0.0354 kg = 3 5 . 4 g [/latex]

Cooling with cold water can be handled the same way. All we need to do is replace the terms for ice by the ones for cold water at 0°C:

[latex] (\Delta U)_{ ext {coldwater }}+(\Delta U)_{ ext {water }}=0 [/latex]

[latex] \left[m C\left(T_{2}-T_{1} ight) ight]_{ ext {coldwater }}+\left[m C\left(T_{2}-T_{1} ight) ight]_{ ext {water }}=0 [/latex]

Substituting,

[latex] \left[m_{ ext {cold water }}\left(4.18 kJ / kg \cdot{ }^{\circ} C ight)(5-0){ }^{\circ} C ight]+(0.2 kg )\left(4.18 kJ / kg \cdot{ }^{\circ} C ight)(5-20)^{\circ} C = 0 [/latex]

It gives [latex] m = 0.6 kg = 600 g [/latex]

Discussion Note that this is 17 times the amount of ice needed, and it explains why we use ice instead of water to cool drinks.

Question 1.132: A 0.2-L glass of water at 20°C is to be cooled with ice to 5...

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