a piece of copper of mass 250gram at a temperature of 900 degree Celsius is quickly transformed to a vessel of negligible thermal capable containing 200 gram of water at 20degree Celsius.if the final temperature is 100degree celsius.what mass of water will boil away?(constant of copper =0.4joulegramkilo, lg=2200joulegram
I need the answer
To solve this problem, we need to calculate the amount of heat lost by the copper and gained by the water. The heat lost by the copper will raise the temperature of the water from 20°C to 100°C and also boil off some mass of water.
First, let's calculate the heat lost by the copper:
Q1 = mass of copper * specific heat of copper * change in temperature
Q1 = 250 g * 0.4 J/g°C * (900°C - 100°C)
= 250 g * 0.4 J/g°C * 800°C
= 80000 J
Now, let's calculate the heat gained by the water:
Q2 = mass of water * specific heat of water * change in temperature
Q2 = 200 g * 4.18 J/g°C * (100°C - 20°C)
= 200 g * 4.18 J/g°C * 80°C
= 66880 J
The heat gained by the water will be used to raise the temperature and boil off some of the water. The heat required to raise the temperature of water is calculated using the specific heat of water, but the heat required to boil water is calculated using the latent heat of vaporization (lg).
Now, let's calculate the amount of water that will boil away:
Q_boiling = mass of water * latent heat of vaporization
Q_boiling = mass of water * lg
Since the final temperature is 100°C, the heat gained by the water is also equal to the heat required to raise the temperature (Q2) plus the heat required for boiling (Q_boiling):
Q2 + Q_boiling = Q1
66880 J + mass of water * lg = 80000 J
mass of water * lg = 80000 J - 66880 J
mass of water * lg = 13120 J
To determine the mass of water that will boil away, we need to divide the heat (13120 J) by the latent heat of vaporization (lg):
mass of water = 13120 J / 2200 J/g
mass of water = 5.96 g
Therefore, approximately 5.96 grams of water will boil away.