Consider the specific heats H2O(s) =2.09 J/g ¡¤◦C,H2O (§¤) = 4.18 J/g ¡¤◦C, and H2O(g) = 2.03 J/g ¡¤◦C. The heat of fusion for water is 334 J/g and its heat of vaporization is 2260 J/g. Calculate the amount of heat required to convert 92 g of ice at −13◦C completely to liquid water at 55◦C.
Answer in units of kJ
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To calculate the amount of heat required to convert 92 g of ice at -13°C to liquid water at 55°C, we need to consider the following steps:
1. Heat absorbed to raise the temperature of ice from -13°C to 0°C.
2. Heat absorbed for the phase change from ice to liquid water at 0°C.
3. Heat absorbed to raise the temperature of liquid water from 0°C to 55°C.
Step 1: Heat absorbed to raise the temperature of ice from -13°C to 0°C
The specific heat of ice (H2O(s)) is 2.09 J/g°C.
The temperature change is 0°C - (-13°C) = 13°C.
Using the formula Q = m * C * ΔT, where Q is the heat absorbed, m is the mass, C is the specific heat, and ΔT is the temperature change:
Q1 = 92 g * 2.09 J/g°C * 13°C
Step 2: Heat absorbed for the phase change from ice to liquid water at 0°C
The heat of fusion for water is 334 J/g.
Using the formula Q = m * ΔHf, where ΔHf is the heat of fusion:
Q2 = 92 g * 334 J/g
Step 3: Heat absorbed to raise the temperature of liquid water from 0°C to 55°C
The specific heat of liquid water (H2O(ℓ)) is 4.18 J/g°C.
The temperature change is 55°C - 0°C = 55°C.
Using the formula Q = m * C * ΔT, where Q is the heat absorbed, m is the mass, C is the specific heat, and ΔT is the temperature change:
Q3 = 92 g * 4.18 J/g°C * 55°C
Finally, to find the total heat required, we add up the heat from all three steps:
Total heat = Q1 + Q2 + Q3
Calculating this expression will give us the answer in units of J. To convert it to kJ, we will divide by 1000.
Let's calculate the result:
Q1 = 92 g * 2.09 J/g°C * 13°C = 2487.4 J
Q2 = 92 g * 334 J/g = 30728 J
Q3 = 92 g * 4.18 J/g°C * 55°C = 209724 J
Total heat = Q1 + Q2 + Q3
Total heat = 2487.4 J + 30728 J + 209724 J = 242939.4 J
Converting to kJ:
Total heat = 242939.4 J / 1000 = 242.9394 kJ
Therefore, the amount of heat required to convert 92 g of ice at -13°C completely to liquid water at 55°C is approximately 242.9394 kJ.
To calculate the amount of heat required to convert 92 g of ice at -13°C to liquid water at 55°C, we need to consider three different processes: heating the ice from -13°C to 0°C, melting the ice at 0°C, and then heating the resulting water from 0°C to 55°C.
Let's break down the calculation step by step:
1. Heating the ice from -13°C to 0°C:
First, calculate the amount of heat required to raise the temperature of the ice using its specific heat:
Q1 = mass × specific heat × temperature change
= 92 g × 2.09 J/g°C × (0°C - (-13°C))
= 92 g × 2.09 J/g°C × 13°C
= 2459.24 J
2. Melting the ice at 0°C:
To melt the ice, we need to calculate the heat of fusion:
Q2 = mass × heat of fusion
= 92 g × 334 J/g
= 30728 J
3. Heating the resulting water from 0°C to 55°C:
Now, calculate the amount of heat required to raise the temperature of the water:
Q3 = mass × specific heat × temperature change
= 92 g × 4.18 J/g°C × (55°C - 0°C)
= 92 g × 4.18 J/g°C × 55°C
= 21731.6 J
Finally, add up the three quantities of heat to get the total heat required:
Total heat = Q1 + Q2 + Q3
= 2459.24 J + 30728 J + 21731.6 J
= 54918.84 J
Since the question asks for the answer in kJ (kilojoules), we divide the total heat by 1000 to convert it to kJ:
Total heat in kJ = 54918.84 J / 1000
= 54.91884 kJ
So, the amount of heat required to convert 92 g of ice at -13°C completely to liquid water at 55°C is approximately 54.92 kJ.