Using calories, calculate how much heat 32.0g of water absorbs when it is heated from 25.0 degrees C to 80.0 degrees C. How many joules is this?
q = mass x specific heat x delta T.
For calories, sp.h. is 1 cal/g*C
For joules, sp.h. is 4.184 J/g*C
To calculate the heat absorbed by the water, you can use the specific heat formula:
Q = m * c * ΔT
where:
Q is the heat absorbed or released by the substance,
m is the mass of the substance,
c is the specific heat capacity of the substance,
ΔT is the change in temperature.
First, let's determine the values:
m (mass of water) = 32.0g
c (specific heat capacity of water) = 1 calorie/gram°C (or 4.184 joules/gram°C)
ΔT (change in temperature) = (final temperature - initial temperature)
ΔT = 80.0°C - 25.0°C = 55.0°C
Now, let's substitute the values into the formula:
Q = 32.0g * 1 calorie/gram°C * 55.0°C
Q = 1,760 calories
To convert calories to joules, we use the conversion factor:
1 calorie = 4.184 joules
So, let's convert the calories into joules:
Q (joules) = 1,760 calories * (4.184 joules/1 calorie)
Q (joules) = 7,354.24 joules
Therefore, 32.0g of water absorbs 1,760 calories or 7,354.24 joules of heat when heated from 25.0°C to 80.0°C.