In a physics experiment, 250g of water is placed in a 50g calorimeter cup made of aluminum when both are at 15C. Steam at 100C is completely absorbed by passing it through the water, raising the temperature of water and cup to 45C. Find the mass of steam absorbed. Specific heat of aluminum is 0.215 cal/gC.
To find the mass of steam absorbed, we need to use the principle of energy conservation. The energy gained by the water and the cup is equal to the energy lost by the steam.
The energy gained by the water and the cup can be calculated using the specific heat formula:
Energy gained = mass * specific heat * change in temperature
First, let's find the energy gained by the water:
mass of water = 250g
specific heat of water = 1 cal/g°C (specific heat of water is 1, as it is given that the values are in calories)
change in temperature = (45 - 15)°C = 30°C
Energy gained by water = mass of water * specific heat of water * change in temperature
Energy gained by water = 250g * 1 cal/g°C * 30°C = 7500 cal
Next, let's find the energy gained by the aluminum cup:
mass of cup = 50g
specific heat of aluminum = 0.215 cal/g°C
change in temperature = (45 - 15)°C = 30°C
Energy gained by cup = mass of cup * specific heat of aluminum * change in temperature
Energy gained by cup = 50g * 0.215 cal/g°C * 30°C = 322.5 cal
Now, since energy is conserved, the energy lost by the steam is equal to the energy gained by the water and the cup:
Energy lost by steam = Energy gained by water + Energy gained by cup
Energy lost by steam = 7500 cal + 322.5 cal = 7822.5 cal
Lastly, let's find the mass of steam absorbed.
To do this, we need to know the specific heat of steam. The specific heat of steam is approximately 0.48 cal/g°C.
Using the formula for energy,
Energy lost by steam = mass of steam * specific heat of steam * change in temperature
Rearranging the formula,
mass of steam = Energy lost by steam / (specific heat of steam * change in temperature)
mass of steam = 7822.5 cal / (0.48 cal/g°C * (100 - 45)°C)
mass of steam = 7822.5 cal / (0.48 cal/g°C * 55°C)
mass of steam ≈ 312.9 g
Therefore, the mass of steam absorbed in the experiment is approximately 312.9 grams.
To solve this problem, we can use the heat equation:
Q = mcΔT,
where Q represents the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the heat transferred to the water:
Q_water = m_water * c_water * ΔT_water,
where m_water is the mass of water and c_water is the specific heat capacity of water.
Given:
m_water = 250g,
c_water = 1 cal/gC (specific heat capacity of water),
ΔT_water = 45C - 15C = 30C.
Substituting the values into the equation:
Q_water = 250g * 1 cal/gC * 30C,
Q_water = 7500 cal.
Now, let's calculate the heat transferred to the aluminum cup:
Q_cup = m_cup * c_cup * ΔT_cup,
where m_cup is the mass of the cup and c_cup is the specific heat capacity of aluminum.
Given:
m_cup = 50g,
c_cup = 0.215 cal/gC (specific heat capacity of aluminum),
ΔT_cup = 45C - 15C = 30C.
Substituting the values into the equation:
Q_cup = 50g * 0.215 cal/gC * 30C,
Q_cup = 322.5 cal.
Since the heat absorbed by the water and cup came from the steam, the total heat absorbed by the system is the sum of the heat absorbed by the water and the heat absorbed by the cup:
Q_total = Q_water + Q_cup,
Q_total = 7500 cal + 322.5 cal,
Q_total = 7822.5 cal.
Next, we need to calculate the heat released by the steam as it condenses into water. The heat released by the steam is equal to the heat absorbed by the water and cup:
Q_steam = Q_total.
Finally, let's calculate the mass of steam absorbed:
Q_steam = m_steam * Hvap,
where Hvap is the heat of vaporization of water.
Given:
Hvap = 540 cal/g (heat of vaporization of water).
Substituting the values into the equation:
7822.5 cal = m_steam * 540 cal/g,
m_steam = 7822.5 cal / 540 cal/g,
m_steam ≈ 14.5 g.
Therefore, the mass of the steam absorbed is approximately 14.5g.