Answer :
To solve this, we can use the formula:
\[ Q = mcΔT \]
Where:
- \( Q \) is the energy in joules
- \( m \) is the mass in grams
- \( c \) is the specific heat capacity of water (4.18 J/g°C)
- \( ΔT \) is the change in temperature in °C
Let's go through each part:
a.) To find the total energy involved, we need to account for both raising the temperature and phase change.
\[ Q_{total} = Q_{raising \ temperature} + Q_{phase \ change} \]
\[ Q_{raising \ temperature} = mcΔT \]
\[ Q_{phase \ change} = mL \]
Where \( L \) is the latent heat of vaporization for water (2260 J/g).
\[ Q_{raising \ temperature} = (240 \ g)(4.18 \ J/g°C)(100°C - T_i) \]
\[ Q_{phase \ change} = (240 \ g)(2260 \ J/g) \]
Given that it takes 16 minutes (960 seconds) to turn the water into steam:
\[ Q_{total} = Q_{raising \ temperature} + Q_{phase \ change} = mcΔT + mL \]
\[ Q_{total} = (240 \ g)(4.18 \ J/g°C)(100°C - T_i) + (240 \ g)(2260 \ J/g) \]
\[ Q_{total} = 240 \ g \times (4.18 \ J/g°C \times 100°C - 4.18 \ J/g°C \times T_i + 2260 \ J/g) \]
\[ Q_{total} = 240 \ g \times (418 \ J - 4.18T_i + 2260 \ J) \]
\[ Q_{total} = 240 \ g \times (2678 - 4.18T_i) \]
\[ Q_{total} = 240 \ g \times 2678 \ J - 1003.2T_i \]
\[ Q_{total} = 642720 \ J - 1003.2T_i \]
Given that the power output of the microwave is 1000 watts and it takes 16 minutes to turn the water into steam:
\[ Q_{total} = P \times t \]
\[ 642720 \ J - 1003.2T_i = 1000 \ W \times 960 \ s \]
\[ 642720 \ J - 1003.2T_i = 960000 \ J \]
\[ -1003.2T_i = 960000 \ J - 642720 \ J \]
\[ -1003.2T_i = 317280 \ J \]
\[ T_i = \frac{317280 \ J}{-1003.2} \]
\[ T_i ≈ -316 \ °C \]
Since a negative temperature is not possible, it means the water was initially frozen and had to be heated up from below 0°C.
b.) The energy involved only in evaporating the water when it is at 100°C is the phase change energy, which is \( mL \), where \( L \) is the latent heat of vaporization for water (2260 J/g).
\[ Q_{phase \ change} = (240 \ g)(2260 \ J/g) \]
\[ Q_{phase \ change} = 240 \ g \times 2260 \ J \]
\[ Q_{phase \ change} ≈ 542400 \ J \]
c.) The energy involved only in raising the water's temperature to its boiling point can be calculated using the formula \( Q_{raising \ temperature} = mcΔT \).
Given that the water's boiling point is 100°C, and its initial temperature is below 0°C, we've already calculated it in part d. Let's use the value we found for \( T_i \).
\[ Q_{raising \ temperature} = (240 \ g)(4.18 \ J/g°C)(100°C - (-316°C)) \]
\[ Q_{raising \ temperature} = (240 \ g)(4.18 \ J/g°C)(416°C) \]
\[ Q_{raising \ temperature} = 401664 \ J \]
So, the amount of energy involved only in raising the water's temperature to its boiling point is approximately 401,664 J.
d.) We've already calculated the initial temperature in part d: \( T_i ≈ -316°C \).
\[ Q = mcΔT \]
Where:
- \( Q \) is the energy in joules
- \( m \) is the mass in grams
- \( c \) is the specific heat capacity of water (4.18 J/g°C)
- \( ΔT \) is the change in temperature in °C
Let's go through each part:
a.) To find the total energy involved, we need to account for both raising the temperature and phase change.
\[ Q_{total} = Q_{raising \ temperature} + Q_{phase \ change} \]
\[ Q_{raising \ temperature} = mcΔT \]
\[ Q_{phase \ change} = mL \]
Where \( L \) is the latent heat of vaporization for water (2260 J/g).
\[ Q_{raising \ temperature} = (240 \ g)(4.18 \ J/g°C)(100°C - T_i) \]
\[ Q_{phase \ change} = (240 \ g)(2260 \ J/g) \]
Given that it takes 16 minutes (960 seconds) to turn the water into steam:
\[ Q_{total} = Q_{raising \ temperature} + Q_{phase \ change} = mcΔT + mL \]
\[ Q_{total} = (240 \ g)(4.18 \ J/g°C)(100°C - T_i) + (240 \ g)(2260 \ J/g) \]
\[ Q_{total} = 240 \ g \times (4.18 \ J/g°C \times 100°C - 4.18 \ J/g°C \times T_i + 2260 \ J/g) \]
\[ Q_{total} = 240 \ g \times (418 \ J - 4.18T_i + 2260 \ J) \]
\[ Q_{total} = 240 \ g \times (2678 - 4.18T_i) \]
\[ Q_{total} = 240 \ g \times 2678 \ J - 1003.2T_i \]
\[ Q_{total} = 642720 \ J - 1003.2T_i \]
Given that the power output of the microwave is 1000 watts and it takes 16 minutes to turn the water into steam:
\[ Q_{total} = P \times t \]
\[ 642720 \ J - 1003.2T_i = 1000 \ W \times 960 \ s \]
\[ 642720 \ J - 1003.2T_i = 960000 \ J \]
\[ -1003.2T_i = 960000 \ J - 642720 \ J \]
\[ -1003.2T_i = 317280 \ J \]
\[ T_i = \frac{317280 \ J}{-1003.2} \]
\[ T_i ≈ -316 \ °C \]
Since a negative temperature is not possible, it means the water was initially frozen and had to be heated up from below 0°C.
b.) The energy involved only in evaporating the water when it is at 100°C is the phase change energy, which is \( mL \), where \( L \) is the latent heat of vaporization for water (2260 J/g).
\[ Q_{phase \ change} = (240 \ g)(2260 \ J/g) \]
\[ Q_{phase \ change} = 240 \ g \times 2260 \ J \]
\[ Q_{phase \ change} ≈ 542400 \ J \]
c.) The energy involved only in raising the water's temperature to its boiling point can be calculated using the formula \( Q_{raising \ temperature} = mcΔT \).
Given that the water's boiling point is 100°C, and its initial temperature is below 0°C, we've already calculated it in part d. Let's use the value we found for \( T_i \).
\[ Q_{raising \ temperature} = (240 \ g)(4.18 \ J/g°C)(100°C - (-316°C)) \]
\[ Q_{raising \ temperature} = (240 \ g)(4.18 \ J/g°C)(416°C) \]
\[ Q_{raising \ temperature} = 401664 \ J \]
So, the amount of energy involved only in raising the water's temperature to its boiling point is approximately 401,664 J.
d.) We've already calculated the initial temperature in part d: \( T_i ≈ -316°C \).
