A) How much energy is released when 25.0 g of steam condense to liquid H2O at 100°C?
B) If it takes 1215J of heat to raise the temperature of water 11.C, how much does the water weigh, in grams?
C) How many joules of heat does it take for 1200.00 grams of water to go from 65C to 98C?
A) heat of vaporization * 25.0 g = ? Joules
B) mass * specific heat * 11ºC = 1215 J
C) 1200.00 g * specific heat * (98ºC - 65ºC) = ? J
you should be aware of significant figure rules in your answers
A) To calculate the amount of energy released when 25.0 g of steam condenses to liquid water at 100°C, we can use the heat of vaporization formula:
Q = m * ΔHvap
Where:
Q = energy released (in joules)
m = mass of the substance (in grams)
ΔHvap = heat of vaporization (in joules/gram)
The heat of vaporization for water is approximately 2.257 J/g.
Q = 25.0 g * 2.257 J/g
Q = 56.425 J
Therefore, approximately 56.425 J of energy is released when 25.0 g of steam condenses to liquid water at 100°C.
B) To calculate the weight of water in grams if it takes 1215 J of heat to raise its temperature by 11.0°C, we can use the specific heat formula:
Q = m * ΔT * C
Where:
Q = energy absorbed (in joules)
m = mass of the substance (in grams)
ΔT = change in temperature (in °C)
C = specific heat capacity (in J/g°C)
The specific heat capacity of water is approximately 4.18 J/g°C.
Q = 1215 J
ΔT = 11.0°C
C = 4.18 J/g°C
1215 J = m * 11.0°C * 4.18 J/g°C
m = 1215 J / (11.0°C * 4.18 J/g°C)
m ≈ 26.077 g
Therefore, the water weighs approximately 26.077 grams.
C) To calculate the amount of heat required for 1200.00 g of water to go from 65°C to 98°C, we can use the specific heat formula:
Q = m * ΔT * C
Where:
Q = energy absorbed (in joules)
m = mass of the substance (in grams)
ΔT = change in temperature (in °C)
C = specific heat capacity (in J/g°C)
The specific heat capacity of water is approximately 4.18 J/g°C.
m = 1200.00 g
ΔT = (98°C - 65°C) = 33°C
C = 4.18 J/g°C
Q = 1200.00 g * 33°C * 4.18 J/g°C
Q ≈ 160,728 J
Therefore, it takes approximately 160,728 J of heat for 1200.00 g of water to go from 65°C to 98°C.
A) To calculate the amount of energy released when steam condenses to liquid water, you need to know the heat of vaporization for water and the mass of steam. The heat of vaporization for water is approximately 2260 J/g.
First, calculate the amount of energy released during condensation using the formula:
Energy released = mass of steam * heat of vaporization
mass of steam = 25.0 g (given)
heat of vaporization = 2260 J/g
Energy released = 25.0 g * 2260 J/g = 56500 J
Therefore, 25.0 g of steam condensing to liquid water at 100°C releases 56500 J of energy.
B) To calculate the weight of water based on the amount of heat required to raise its temperature, you need to know the specific heat capacity of water and the amount of heat. The specific heat capacity of water is 4.18 J/g°C.
First, calculate the weight of water using the formula:
Heat = mass of water * specific heat capacity * change in temperature
mass of water = ?
specific heat capacity = 4.18 J/g°C (given)
change in temperature = 11°C (given)
heat = 1215 J (given)
Rearrange the formula to solve for mass of water:
mass of water = heat / (specific heat capacity * change in temperature)
mass of water = 1215 J / (4.18 J/g°C * 11°C) = 25 g
Therefore, the water weighs 25 grams.
C) To calculate the amount of heat required to increase the temperature of water, you need to know the specific heat capacity of water, the mass of water, and the change in temperature.
First, calculate the amount of heat using the formula:
Heat = mass of water * specific heat capacity * change in temperature
mass of water = 1200.00 g (given)
specific heat capacity = 4.18 J/g°C (given)
change in temperature = 98°C - 65°C = 33°C
Heat = 1200.00 g * 4.18 J/g°C * 33°C = 166392 J
Therefore, it takes 166392 J of heat for 1200.00 grams of water to go from 65°C to 98°C.