The specific heat capacity of ice is about 0.5 cal/g degree C
A)Supposing that it remains at that value all the way to absolute zero, calculate the number of calories it would take to change a 0.50g ice cube at absolute zero (-273 degree C}) to 0.50 g of boiling water.
B)How does this number of calories compare with the number of calories required to change the same gram of 100 degree C boiling water to 100 degree C steam?
To calculate the number of calories required to change a 0.50g ice cube at absolute zero (-273 degree C) to 0.50g of boiling water, we need to consider two processes: raising the temperature of the ice to its melting point, and then changing the state of the water from solid to liquid by melting it.
A) Raising the temperature of the ice:
The specific heat capacity of ice is given as 0.5 cal/g degree C. Since the ice is currently at absolute zero (-273 degree C), we need to raise its temperature to its melting point (0 degree C).
The temperature change required is ∆T = 0 degree C - (-273 degree C) = 273 degree C.
The heat (Q) required to raise the temperature of the ice can be calculated using the formula:
Q = mass (m) x specific heat capacity (c) x ∆T
Substituting the given values:
Q = 0.50g x 0.5 cal/g degree C x 273 degree C
Q = 68.25 calories
So, it would take 68.25 calories to raise the temperature of the 0.50g ice cube from absolute zero (-273 degree C) to its melting point (0 degree C).
B) Melting the ice:
To change the state of the ice from solid to liquid (melting), we need to consider the heat of fusion. The heat of fusion for ice is 79.7 cal/g.
The heat (Q) required to melt the ice can be calculated using the formula:
Q = mass (m) x heat of fusion (Hf)
Substituting the given values:
Q = 0.50g x 79.7 cal/g
Q = 39.85 calories
So, it would take 39.85 calories to melt the 0.50g ice cube.
Adding the two calculations together, we get the total number of calories required:
Total Q = 68.25 calories + 39.85 calories
Total Q ≈ 108.1 calories
Therefore, it would take approximately 108.1 calories to change a 0.50g ice cube at absolute zero to 0.50g of boiling water.
B) Comparing with the number of calories required to change the same gram of boiling water to steam:
To change the state of boiling water at 100 degree C to steam at 100 degree C, we need to consider the heat of vaporization. The heat of vaporization for water is 540 cal/g.
Using the same formula as above:
Q = mass (m) x heat of vaporization (Hv)
Substituting the given values:
Q = 0.50g x 540 cal/g
Q = 270 calories
Therefore, it would take approximately 270 calories to change the same gram of boiling water to steam.
Comparing the two calculations:
To change the gram of boiling water to steam requires 270 calories, while changing the gram of ice at absolute zero to boiling water requires approximately 108.1 calories.
Hence, it requires less energy to change the gram of ice at absolute zero to boiling water compared to changing the gram of boiling water to steam.