Calculate the joules of energy involved when 11 grams of water at 10.0 degrees Celsius cools and then freezes at 0.0 degrees Celsius.
How many kilocalories are required to raise the temperature of 305 gtams silver from 30 degrees Celsius to 70 degrees Celsius? (spec. heat of Ag =0.0g cal/g degrees Celsius)
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q1 = heat released to cool water from 10 C to zero C.
q1 = mass water x specific heat water x (Tfinal-Tinitial).
q2 = heat released to freeze liquid water at zero C to solid ice at zero C.
qw2 = mass water x heat fusion.
Total q = q1 + q2
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To calculate the joules of energy involved when water cools and freezes, we need to consider two separate steps: cooling from 10.0 degrees Celsius to 0.0 degrees Celsius, and then freezing at 0.0 degrees Celsius.
1. Cooling from 10.0 degrees Celsius to 0.0 degrees Celsius:
The specific heat capacity of water is approximately 4.18 J/g°C. We can use the formula Q = mcΔT, where Q is the heat transferred, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
Mass of water (m) = 11 grams
Specific heat capacity of water (c) = 4.18 J/g°C
Change in temperature (ΔT) = (0.0°C - 10.0°C) = -10.0°C
Using the formula Q = mcΔT, we have:
Q = (11 g) × (4.18 J/g°C) × (-10.0°C)
Q = -459.8 J
2. Freezing at 0.0 degrees Celsius:
The heat of fusion for water is approximately 334 J/g. The heat of fusion is the amount of energy required for a substance to change from a liquid to a solid state without changing its temperature. Again, we can use the formula Q = mcΔT, where Q is the heat transferred, m is the mass of water, and ΔT is the change in temperature. In this case, the change in temperature is zero.
Given:
Mass of water (m) = 11 grams
Heat of fusion for water (c) = 334 J/g
Using the formula Q = mcΔT, we have:
Q = (11 g) × (334 J/g) × (0.0°C - 0.0°C)
Q = 0 J
So, the total energy involved when 11 grams of water at 10.0 degrees Celsius cools and freezes at 0.0 degrees Celsius is -459.8 J.
For the second question, to calculate the kilocalories required to raise the temperature of silver from 30 degrees Celsius to 70 degrees Celsius, we use the formula Q = mcΔT , where Q is the heat transferred, m is the mass of silver, c is the specific heat capacity of silver, and ΔT is the change in temperature.
Given:
Mass of silver (m) = 305 grams
Specific heat capacity of silver (c) = 0.0 cal/g°C
Change in temperature (ΔT) = (70°C - 30°C) = 40°C
Using the formula Q = mcΔT, we have:
Q = (305 g) × (0.0 cal/g°C) × (40°C)
Q = 0 cal
Since we have the result in calories (cal), we can convert it to kilocalories (kcal) by dividing by 1000:
0 cal ÷ 1000 = 0 kcal
Therefore, the number of kilocalories required to raise the temperature of 305 grams of silver from 30 degrees Celsius to 70 degrees Celsius is 0 kcal.
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