How much energy does it take to convert 750 grams of ice at 0 degrees Celsius to liquid water at 40 degrees Celsius? (Specific Heat of water = 4.18 J/g*C and the heat of fusion = 334 J/g)
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To calculate the amount of energy required to convert ice at 0 degrees Celsius to liquid water at 40 degrees Celsius, we need to consider two separate processes: the energy required to melt the ice (heat of fusion) and the energy required to heat the liquid water (specific heat).
First, let's calculate the energy required to melt the ice:
1. Calculate the heat of fusion:
The heat of fusion is given as 334 J/g.
Energy = mass × heat of fusion
Energy = 750 g × 334 J/g
Energy = 250,500 J
Therefore, it takes 250,500 Joules (J) of energy to convert 750 grams of ice to water at 0 degrees Celsius.
Next, let's calculate the energy required to heat the liquid water:
2. Calculate the temperature change:
The initial temperature of the liquid water is 0 degrees Celsius, and the final temperature is 40 degrees Celsius.
Temperature change = final temperature - initial temperature
Temperature change = 40°C - 0°C
Temperature change = 40°C
3. Calculate the energy required to heat the liquid water:
The specific heat of water is given as 4.18 J/g°C
Energy = mass × specific heat × temperature change
Energy = 750 g × 4.18 J/g°C × 40°C
Energy = 125,400 J
Therefore, it takes 125,400 Joules (J) of energy to heat 750 grams of water from 0 to 40 degrees Celsius.
Finally, we add the energy required for melting the ice to the energy required for heating the liquid water:
Total energy = energy to melt ice + energy to heat water
Total energy = 250,500 J + 125,400 J
Total energy = 375,900 J
Therefore, it takes a total of 375,900 Joules (J) of energy to convert 750 grams of ice at 0 degrees Celsius to liquid water at 40 degrees Celsius.
To melt the ice is q1 = mass ice x heat fusion.
To raise T from zero to 40 C is
q2 = mass water x specific heat water x 40
Total is sum q1 + q2.