High School

How many kilojoules of energy are needed to convert 98.2 g of ice at -14.8°C to water at 21.9°C?

(Note: The specific heat of ice is 2.01 J/g°C.)

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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