An ice-cube tray of negligible mass contains 2.5 kg of water at 25°C. How much heat must be removed to cool the water to 0°C and freeze it?

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madeleinejames20 · Accepted Answer

To solve this problem, we need to consider two steps: cooling the water from 25°C to 0°C and then freezing it at 0°C.Let&#039;s begin with the first step, cooling the water from 25°C to 0°C. The amount of heat required to cool the water can be calculated using the specific heat capacity formula:Q1=mcΔTWhere:- Q1 is the amount of heat required- m is the mass of the water- c is the specific heat capacity of water- ΔT is the change in temperatureThe specific heat capacity of water is approximately 4.18J/g°C.Given that the mass of the water is 2.5 kg, we need to convert it to grams:m=2.5kg×1000g/kg=2500gThe change in temperature is:ΔT=0°C−25°C=−25°CNow, we can substitute the values into the formula:Q1=2500g×4.18J/g°C×(−25°C)Simplifying the equation gives us:Q1=−261,250JThe negative sign indicates that heat is being removed from the water.Moving on to the second step, freezing the water at 0°C. To freeze water, we need to remove the heat of fusion, which is the amount of heat required to change the state of a substance from a liquid to a solid. The heat of fusion for water is 334J/g.Using the same mass of water (2500 g), we can calculate the amount of heat required to freeze the water:Q2=2500g×334J/gSimplifying the equation gives us:Q2=835,000JNow, to find the total amount of heat required to cool the water to 0°C and freeze it, we add the heat required for both steps:Qtotal=Q1+Q2Qtotal=−261,250J+835,000JQtotal=573,750JTherefore, to cool the water to 0°C and freeze it, approximately 573,750 Joules of heat must be removed.

An ice-cube tray of negligible mass contains 2.5

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