Calculate the amount of heat required to convert 15 gm water at 100
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To calculate the amount of heat required to convert 15 grams of water at 100°C, we need to consider two separate steps: raising the temperature of the water from its initial temperature to the boiling point, and then converting the water into steam.
To calculate the heat required for the first step, we'll use the specific heat capacity of water. The specific heat capacity is the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius per unit mass. For water, the specific heat capacity is approximately 4.18 J/g°C.
Step 1: Raising the temperature of water to the boiling point
The initial temperature is assumed to be 25°C (room temperature) for this example. The change in temperature required is 100°C - 25°C = 75°C.
The heat energy required (Q1) can be calculated using the formula:
Q1 = mass × specific heat capacity × change in temperature
Plugging in the values:
Q1 = 15 g × 4.18 J/g°C × 75°C = 4702.5 J
So, the heat required to raise the temperature of the water to the boiling point is 4702.5 joules.
Step 2: Converting the water into steam
To convert water at its boiling point into steam, we need to consider the heat of vaporization. The heat of vaporization for water is approximately 2260 J/g.
The heat energy required (Q2) can be calculated using the formula:
Q2 = mass × heat of vaporization
Plugging in the values:
Q2 = 15 g × 2260 J/g = 33,900 J
So, the heat required to convert the water into steam is 33,900 joules.
Finally, to calculate the total heat required, we add the heat required for both steps:
Total heat = Q1 + Q2
Total heat = 4702.5 J + 33,900 J = 38,602.5 J
Therefore, the total heat required to convert 15 grams of water at 100°C into steam is approximately 38,602.5 joules.