calculate the amount of heat (kcal) released when 50.0 g of water at 100degrees Celsius hits the skin and cools to a body temperature of 37 degrees Celsius.
Well, well, well, looks like we have a toasty situation here. Let's do some heat shenanigans!
First, we need to find the heat absorbed when water cools from 100°C to 37°C. And to do that, we'll use the formula:
q = m * C * ΔT
Where:
q is the heat absorbed/released (in calories),
m is the mass (in grams),
C is the specific heat of water (1 cal/g°C),
ΔT is the change in temperature (final temp - initial temp).
Okay, let's plug in the values and see what pops out:
q = 50.0 g * 1 cal/g°C * (37°C - 100°C)
Now, you might be wondering why we didn't use the body temperature of 37°C to start with. That's because initially, we need to consider the water at its boiling point of 100°C.
Anyway, let's continue:
q = 50.0 g * 1 cal/g°C * (-63°C)
Calculating that, you get:
q = -3150 cal
Oh my, it seems like we have some negative heat. But fear not, my friend! It just means heat is being released, not absorbed. So, the amount of heat released when 50.0 g of water cools from 100°C to 37°C is -3150 calories or -3.15 kilocalories (since there are 1000 calories in a kilocalorie).
Now, remember that this calculation assumes no heat loss to the surroundings, and it's purely hypothetical. Stay cool, my friend!
To calculate the amount of heat released, we can use the equation:
Q = mcΔT
Where:
Q is the amount of heat in calories (cal) or kilocalories (kcal)
m is the mass of the substance in grams (g)
c is the specific heat capacity of the substance in cal/g°C (calories per gram per degree Celsius)
ΔT is the change in temperature in degrees Celsius (°C)
First, let's find the specific heat capacity of water. The specific heat capacity of water is 1 cal/g°C.
Next, we can calculate the change in temperature (ΔT) of the water:
ΔT = final temperature - initial temperature
ΔT = 37°C - 100°C
ΔT = -63°C
Since water is cooling down, the change in temperature will be negative.
Now, we can substitute the values into the equation:
Q = mcΔT
Q = (50.0 g) * (1 cal/g°C) * (-63°C)
Q = -3150 cal
Finally, let's convert the amount of heat to kilocalories (kcal), which is the more commonly used unit. We divide the result by 1000:
Q = (-3150 cal) / 1000
Q = -3.15 kcal
Therefore, when 50.0 g of water at 100°C hits the skin and cools to a body temperature of 37°C, approximately 3.15 kcal of heat are released. Remember, negative sign indicates that heat is released.