What volume of water at 0°C can a freezer make into ice cubes in 1.0 hour,if the coefficient of performance of the cooling unit is 7.0 and the power input is 1.0 kilowatt?

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What volume of water at 0°C can a freezer make into ice cubes in 1.0 hour,if the coefficient of performance of the cooling unit is 7.0 and the power input is 1.0 kilowatt?

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Answer 1 · 2018-03-12T14:57:32

$8.82 L$ $8.82 L$

Explanation:

Power input of the refrigerator is $1 k W$ $1 kW$ means per second is supplies $1 \cdot 10^{3} J$ $1*10^3J$ of energy

So,in $1 h r$ $1 hr$ it will supply $10^{3} \cdot 3600 = 36 \cdot 10^{5} J = 8.56 \cdot 10^{5}$ $10^3 *3600=36*10^5 J=8.56*10^5$ calories of heat energy.

Now,let the refrigerator will take the temperature of water to $T_{2}$ $T_2$ and the water is at $T_{1}$ $T_1$ temperature.

Given, $β = 7, T_{1} = 0 + 273 = 273 K$ $beta =7,T_1=0+273=273K$

So,using the equation, $β = \frac{T_{2}}{T_{1} - T_{2}}$ $beta =(T_2)/(T_1 -T_2)$

We,get, $T_{2} = 238.875 K = - {34.125}^{\circ} C$ $T_2=238.875K=-34.125^@C$

Now,heat energy required for conversion of $m$ $m$ $g r a m$ $gram$ of water at $0^{\circ} C$ $0^@C$ to same amount of ice at $0^{\circ} C$ $0^@C$ i.e the latent heat required is $80 m$ $80m$ (as,latent heat of water is $80$ $80$ calories per gram)

So,to take this $m$ $m$ $g r a m$ $gram$ of ice at $0^{\circ} C$ $0^@C$ to $- {34.125}^{\circ} C$ $-34.125^@C$ heat energy required will be $m \cdot 0.5 \cdot (0 - (- 34.125))$ $m*0.5*(0-(-34.125))$ calories (using the formula, heat required $H = m s d θ$ $H=ms d theta$ ,where, $s$ $s$ is the specific heat,for ice it is $0.5$ $0.5$ C.G.S units and $d θ$ $d theta$ is the change in temperature)

So,this entire amount of heat will be equals to $8.56 \cdot 10^{5}$ $8.56*10^5$ calories

So, $80 m + m \cdot 0.5 \cdot 34.125 = 8.56 \cdot 10^{5}$ $80m+m*0.5*34.125=8.56*10^5$

we get,from the above equation, $m = 8819.06$ $m=8819.06$ $g r a m$ $gram$

Now,density of water is $1$ $1$ $\frac{g r a m}{c m^{3}}$ $(gram)/(cm^3)$

So,volume of this much water will be $\frac{8819.06}{1} = 8819.06 c m^{3}$ $8819.06/1=8819.06 cm^3$ (as, $v = \frac{m}{d}$ $v=m/d$ )

Now, $1 c m^{3} = 0.001 L$ $1 cm^3=0.001 L$

so, $8819.06 c m^{3} = 8.81906 = 8.82 L$ $8819.06 cm^3=8.81906=8.82L$

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What volume of water at 0°C can a freezer make into ice cubes in 1.0 hour,if the coefficient of performance of the cooling unit is 7.0 and the power input is 1.0 kilowatt?

1 Answer

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