the temperature of water in a 100 mL container was raised from 26.7 degrees C to 62.3 degrees C. how much heat in calories was added
q = mass water x specific heat water x (Tfinal-Tinitial)
ok i don't really understand the answer to this question.
hit dat dab
To calculate the amount of heat added to the water, you can use the formula:
Q = mcΔT
Where:
Q is the heat added (in calories)
m is the mass of the water (in grams)
c is the specific heat capacity of water (approximately 1 calorie/gram°C)
ΔT is the change in temperature (final temperature - initial temperature)
In this case, we have a 100 mL container of water, which we can assume has a mass of 100 grams since the density of water is approximately 1 gram/mL. The specific heat capacity of water is approximately 1 calorie/gram°C.
Now let's calculate the heat added:
m = 100 grams
c = 1 calorie/gram°C
ΔT = 62.3°C - 26.7°C = 35.6°C
Q = (100 grams) * (1 calorie/gram°C) * (35.6°C)
Simplifying the equation, we get:
Q = 3560 calories
Therefore, approximately 3560 calories of heat was added to raise the temperature of the water in the 100 mL container from 26.7°C to 62.3°C.