140 Cal to heat 355 gm of water with initial temperature of 25 degree celsius, final temp?
140=355*1cal/gram*(t-25)
solve for Temp t
It's 140 Kcal. I got Tf 419.4 deg C. This seemed very high and I can't figure out error
it wont go that high, water boils.
your calculations are correct, but something is wrong with the problem statement, either in the kcal, or the mass.
It's a can of coke problem.... 355 ml of water (assume coke=water)140 Cal in the can. Using energy from sugar in can. What final temp would be reached if initial was 25 deg C. ?
so my question is Cal. is that calories, or Calories (ie, Food Calories=1 kcal).
a) if it is food, then the water heats up to 100C, then boils , and if all of it boils away, the residual heat heats the steam. This is harder. You have to figure the heat of vaporization at 100C, and the specific heat of steam.
If it is calories, then it is the way I suggested.
I think it's a crappy question, a and b section of introduction question all talk about teaspoons of sucrose and calories associated. That leads one to believe we are dealing with nutritional Calorie = 1000 calories right? I don't think it's a heat of vaporization cause we haven't covered that yet. only on ch 6 in open stax text. Any thoughts
To calculate the final temperature of the water, we can use the formula:
Q = mcΔT
Where:
Q is the heat energy absorbed or released by the water (in calories),
m is the mass of the water (in grams),
c is the specific heat capacity of water (which is approximately 1 calorie/gram°C), and
ΔT is the change in temperature (final temperature - initial temperature).
In this case, we know:
Q = 140 calories
m = 355 grams
c = 1 calorie/gram°C
initial temperature = 25°C
We need to find the final temperature (ΔT). Rearranging the formula, we have:
ΔT = Q / (mc)
Plugging in the values, we get:
ΔT = 140 calories / (355 grams * 1 calorie/gram°C)
Simplifying the units, we get:
ΔT = 140 / 355 °C
Calculating this, we find:
ΔT ≈ 0.394°C
To find the final temperature, we add the change in temperature (ΔT) to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 25°C + 0.394°C
Calculating this, we find:
Final temperature ≈ 25.394°C
Therefore, the final temperature of the water is approximately 25.394°C.