An ice bag containing 0°C ice is much more effective in absorbing heat than one containing the same amount of 0°C water.
(a) How much heat in kcal is required to raise the temperature of 0.700 kg of water from 0°C to 27.0°C?
(b) How much heat is required to first melt 0.700 kg of 0°C ice and then raise its temperature?
(c) Explain how your answers support the contention that the ice is more effective.
(a) To calculate the heat required to raise the temperature of water, we can use the specific heat capacity formula:
Q = m * c * ΔT
Where:
Q is the heat energy,
m is the mass of the substance (in kg),
c is the specific heat capacity of the substance (in kcal/kg°C),
ΔT is the change in temperature (in °C).
Given:
m = 0.700 kg
ΔT = 27.0°C - 0°C = 27.0°C
c (water) = 1.00 kcal/kg°C
Using the formula, we can calculate:
Q = 0.700 kg * 1.00 kcal/kg°C * 27.0°C
Q = 18.9 kcal
Therefore, the heat required to raise the temperature of 0.700 kg of water from 0°C to 27.0°C is 18.9 kcal.
(b) To calculate the total heat required to first melt the ice and then raise its temperature, we need to consider two steps.
Step 1: Melting the ice.
To calculate the heat required to melt ice, we can use the formula:
Q1 = m * Lf
Where:
Q1 is the heat required to melt the ice,
m is the mass of the ice (in kg),
Lf is the latent heat of fusion of ice (in kcal/kg).
Given:
m = 0.700 kg
Lf (ice) = 79.7 kcal/kg (latent heat of fusion of ice)
Using the formula, we can calculate:
Q1 = 0.700 kg * 79.7 kcal/kg
Q1 = 55.79 kcal
Step 2: Raising the temperature of water.
To calculate the heat required to raise the temperature of water, we can use the specific heat capacity formula as mentioned in part (a):
Q2 = m * c * ΔT
Given:
m = 0.700 kg
ΔT = 27.0°C - 0°C = 27.0°C
c (water) = 1.00 kcal/kg°C
Using the formula, we can calculate:
Q2 = 0.700 kg * 1.00 kcal/kg°C * 27.0°C
Q2 = 18.9 kcal
Therefore, the total heat required to first melt 0.700 kg of ice and then raise its temperature is:
Total heat = Q1 + Q2 = 55.79 kcal + 18.9 kcal = 74.69 kcal
(c) The fact that it takes more heat to melt the ice and raise its temperature compared to just raising the temperature of the same amount of water supports the contention that the ice is more effective in absorbing heat. This is because the ice undergoes a phase change from solid to liquid (melting) before it can absorb further heat to increase its temperature. This requires additional energy (latent heat of fusion) to break the intermolecular bonds holding the ice molecules together. In contrast, the water starts in the liquid phase, so there is no phase change involved, and it only needs to absorb heat to raise its temperature. Therefore, ice is more effective at absorbing heat as it requires additional energy for the phase change, making it a better cooling agent.
(a) To calculate the amount of heat required to raise the temperature of water, we can use the specific heat capacity equation:
Q = mcΔT
Where:
Q is the heat energy in calories,
m is the mass of the substance in grams,
c is the specific heat capacity of the substance in calories per gram-degree Celsius (cal/g°C), and
ΔT is the change in temperature in degrees Celsius.
In this case, we have:
mass (m) = 0.700 kg = 700 g
specific heat capacity (c) = 1 cal/g°C (for water)
change in temperature (ΔT) = 27°C - 0°C = 27°C
Therefore, the equation becomes:
Q = (700 g) (1 cal/g°C) (27°C)
Q = 18,900 cal
To convert calories to kilocalories, we divide by 1000:
Q = 18,900 cal / 1000 cal/kcal = 18.9 kcal
So, it would require 18.9 kcal of heat to raise the temperature of 0.700 kg of water from 0°C to 27.0°C.
(b) To calculate the heat required to melt the ice and then raise its temperature, we need to consider two processes separately: the heat of fusion and the heat needed to raise the temperature.
The heat of fusion is the amount of heat required to convert a substance from a solid to a liquid state without any change in temperature. For water, the heat of fusion is 79.7 cal/g.
So the amount of heat required to melt 0.700 kg (or 700 g) of ice is:
Q1 = (700 g) (79.7 cal/g)
Q1 = 55,790 cal
To convert calories to kilocalories, we divide by 1000:
Q1 = 55,790 cal / 1000 cal/kcal = 55.79 kcal
After the ice is completely melted, we then need to raise its temperature from 0°C to 27°C. We can use the specific heat capacity of water, which is 1 cal/g°C.
So the amount of heat required to raise the temperature of 0.700 kg (or 700 g) of water from 0°C to 27°C is:
Q2 = (700 g) (1 cal/g°C) (27°C)
Q2 = 18,900 cal
To convert calories to kilocalories, we divide by 1000:
Q2 = 18,900 cal / 1000 cal/kcal = 18.9 kcal
Therefore, the total amount of heat required to first melt 0.700 kg of ice and then raise its temperature is the sum of Q1 and Q2:
Total heat = Q1 + Q2 = 55.79 kcal + 18.9 kcal = 74.69 kcal
(c) The answer to part (b) demonstrates that it requires more total heat (74.69 kcal) to first melt the ice and then raise its temperature compared to just raising the temperature of the same amount of water (18.9 kcal in part a).
This supports the contention that an ice bag containing 0°C ice is more effective in absorbing heat than one containing the same amount of 0°C water. The reason for this is that ice has a higher latent heat of fusion compared to the specific heat capacity of water. This means that ice can absorb a large amount of heat when it melts, without experiencing a significant change in temperature. Therefore, when an ice bag is placed on something warm, it absorbs more heat during the melting process compared to the same mass of water. This makes the ice bag more effective in cooling or absorbing heat from the surroundings.