How many Calories of energy does it take to heat 257 mL of water by 62.2 C?
q = mass H2O x specific heat water x 62.2.
To calculate the number of calories of energy required to heat water, you can use the following formula:
Q = m * C * ΔT
Where:
Q = amount of heat energy required (in calories)
m = mass of water (in grams)
C = specific heat capacity of water (1 calorie/gram °C)
ΔT = change in temperature (in °C)
Given:
Volume of water (V) = 257 mL = 257 grams (since the density of water is approximately 1 g/mL)
ΔT = 62.2 °C
First, you need to convert the volume (in mL) to mass (in grams):
m = V = 257 grams
Now, substitute the values into the formula:
Q = m * C * ΔT
Q = 257 g * 1 calorie/gram °C * 62.2 °C
Calculating the product:
Q = 16,033.4 calories
Therefore, it will take approximately 16,033.4 calories of energy to heat 257 mL of water by 62.2 °C.
To calculate the number of calories required to heat a given volume of water, you can use the formula:
Q = m * C * ΔT
Where:
Q is the heat energy measured in calories,
m is the mass of the substance (water in this case) measured in grams,
C is the specific heat capacity of the substance measured in calories per gram per degree Celsius (cal/g°C),
ΔT is the change in temperature in degrees Celsius.
In this case, we need to heat 257 mL of water by 62.2°C.
To convert 257 mL to grams of water, we need to know the density of water, which is approximately 1 gram per milliliter. Hence, 257 mL of water is equal to 257 grams.
Assuming the specific heat capacity of water is approximately 1 calorie per gram per degree Celsius (cal/g°C), we can now calculate the amount of heat energy required.
Q = m * C * ΔT
Q = 257 g * 1 cal/g°C * 62.2°C
By multiplying these values, we get:
Q = 16,044.6 calories
Therefore, it would take approximately 16,044.6 calories of energy to heat 257 mL of water by 62.2°C.