Answer
(a) First, we need to calculate the amount of heat released by the lava bomb as it cools from 984°C to its melting point and solidifies. This heat will be absorbed by the water in the pool, causing its temperature to rise. The heat released by the lava bomb is given by:
Q1 = (mass of lava bomb) x (specific heat of solid lava) x (984°C - melting point)
Q1 = (15.4 Mg) x (1.04 kJ/kg·K) x (984°C - 1200°C)
Q1 = -2.86 × $10^8 $kJ
The negative sign indicates that heat is being released by the lava bomb.
The amount of heat absorbed by the water in the pool is given by:
Q2 = (mass of water) x (specific heat of water) x (final temperature - initial temperature)
Q2 = (20.4 Mg) x (4.18 kJ/kg·K) x (T - 24.0°C)
where T is the final temperature of the water.
Since the system is in thermal equilibrium, we can equate Q1 and Q2:
-2.86 × $10^8$ kJ = (20.4 Mg) x (4.18 kJ/kg·K) x (T - 24.0°C)
Solving for T, we get:
T = 100.2°C
Therefore, the pool water will reach the boiling point.
(b) Once the pool water reaches the boiling point, it will start to boil and turn into steam. The heat required for this process is given by the heat of vaporization of water, which is 2.26 MJ/kg. The amount of heat required to vaporize all the water in the pool is:
Q3 = (mass of water) x (heat of vaporization of water)
Q3 = (20.4 Mg) x (2.26 MJ/kg)
Q3 = 46.1 × $10^6 $kJ
This is much larger than the heat released by the lava bomb, so the pool water will not boil completely away. Only a fraction of the water will turn into steam, and the rest will remain in the pool, now heated to 100.2°C.
Work Step by Step
(a) First, we need to calculate the amount of heat released by the lava bomb as it cools from 984°C to its melting point and solidifies. This heat will be absorbed by the water in the pool, causing its temperature to rise. The heat released by the lava bomb is given by:
Q1 = (mass of lava bomb) x (specific heat of solid lava) x (984°C - melting point)
Q1 = (15.4 Mg) x (1.04 kJ/kg·K) x (984°C - 1200°C)
Q1 = -2.86 × $10^8 $kJ
The negative sign indicates that heat is being released by the lava bomb.
The amount of heat absorbed by the water in the pool is given by:
Q2 = (mass of water) x (specific heat of water) x (final temperature - initial temperature)
Q2 = (20.4 Mg) x (4.18 kJ/kg·K) x (T - 24.0°C)
where T is the final temperature of the water.
Since the system is in thermal equilibrium, we can equate Q1 and Q2:
-2.86 × $10^8 $kJ = (20.4 Mg) x (4.18 kJ/kg·K) x (T - 24.0°C)
Solving for T, we get:
T = 100.2°C
Therefore, the pool water will reach the boiling point.
(b) Once the pool water reaches the boiling point, it will start to boil and turn into steam. The heat required for this process is given by the heat of vaporization of water, which is 2.26 MJ/kg. The amount of heat required to vaporize all the water in the pool is:
Q3 = (mass of water) x (heat of vaporization of water)
Q3 = (20.4 Mg) x (2.26 MJ/kg)
Q3 = 46.1 × $10^6 $kJ
This is much larger than the heat released by the lava bomb, so the pool water will not boil completely away. Only a fraction of the water will turn into steam, and the rest will remain in the pool, now heated to 100.2°C.
