Answer :
It takes approximately 35.764 kilojoules (kJ) of energy to convert 98.2 grams (g) of ice at -14.8°C to water at 21.9°C.
To find the amount of energy needed to convert ice at -14.8°C to water at 21.9°C, we need to consider two steps:
Step 1: Calculate the energy needed to raise the temperature of ice to 0°C.
The specific heat capacity of ice at -14.8°C is given as 2.01 J/gºC. The temperature change is from -14.8°C to 0°C, which is a difference of 14.8°C.
Using the formula Q = m * c * ΔT, where Q is the energy, m is the mass, c is the specific heat capacity, and ΔT is the temperature change, we can calculate the energy:
Q1 = 98.2 g * 2.01 J/gºC * 14.8ºC
= 2901.528 J
Step 2: Calculate the energy needed to melt the ice.
The heat of fusion for water is 334 J/g.
Using the formula Q = m * ΔHf, where Q is the energy, m is the mass, and ΔHf is the heat of fusion, we can calculate the energy:
Q2 = 98.2 g * 334 J/g
= 32862.8 J
Now, we can add the energies from both steps to find the total energy needed:
Total energy = Q1 + Q2
= 2901.528 J + 32862.8 J
= 35764.328 J
To convert this energy to kilojoules, we divide by 1000:
Total energy in kJ = 35764.328 J / 1000
= 35.764 kJ
Therefore, it takes approximately 35.764 kilojoules (kJ) of energy to convert 98.2 grams (g) of ice at -14.8°C to water at 21.9°C.
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