A 580mL sample of water was cooled from 60°C to 10°C. how much heat was lost?
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Post your work if you get stuck. specific heat H2O = 4.184 J/g*C
Well, let's see. Water, as we all know, has a mind of its own and loves to play hide-and-seek with heat. So, when your 580mL sample of water went from 60°C to 10°C, it definitely had a fun time losing some heat along the way.
Now, to determine how much heat was lost, we need to use the equation Q = mcΔT, where Q represents the heat lost, m is the mass of the water in grams, c is the specific heat capacity of water, and ΔT is the change in temperature.
Since you provided the volume, not the mass, we need to convert it. Considering that the density of water is about 1g/mL, your sample of water weighs roughly... 580 grams! Ta da!
Now, let's take the specific heat capacity of water into account. It's approximately 4.18 J/g°C. And the change in temperature is 60°C - 10°C = 50°C.
Plugging these values into the equation, we get Q = (580g)(4.18 J/g°C)(50°C). Let me just crunch some numbers here... *beep boop beep* ...and the answer is approximately 121,820 Joules.
So, your 580mL sample of water lost about 121,820 Joules of heat during its cool-down adventure. That's quite a chilly journey, if you ask me!
To calculate the amount of heat lost, we can use the formula:
Q = m × c × ΔT
Where:
Q is the amount of heat lost
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
Given:
m = 580 mL of water
c = 4.18 J/g·°C (specific heat capacity of water)
ΔT = 60°C - 10°C = 50°C
First, we need to convert the volume of water from milliliters (mL) to grams (g), assuming the density of water is 1 g/mL.
1 mL = 1 g
So, the mass of the water (m) is 580 g.
Now, we can substitute the values into the formula:
Q = 580 g × 4.18 J/g·°C × 50°C
Calculating:
Q = 121,420 J
Therefore, the amount of heat lost is 121,420 Joules.
To find the amount of heat lost, we need to use the formula:
Q = m * C * ΔT
Where:
Q is the amount of heat lost
m is the mass of the substance (in grams)
C is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (in °C)
First, we need to convert the volume of water (580 mL) to mass (grams). We can do this by multiplying the volume by the density of water.
The density of water is approximately 1 g/mL. So, 580 mL of water is equivalent to 580 grams.
Next, we can calculate the change in temperature (ΔT) by subtracting the final temperature (10°C) from the initial temperature (60°C).
ΔT = 60°C - 10°C = 50°C
Now, we can substitute these values into the formula to calculate the amount of heat lost (Q):
Q = 580 g * 4.18 J/g°C * 50°C
Q = 121,220 J
Therefore, the amount of heat lost is 121,220 Joules.