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To determine the final temperature of a 1,468.7 g sample of water after it has absorbed 19.89 kJ of heat, follow these steps:
1. Convert the heat absorbed into Joules:
Since the heat absorbed is given in kilojoules (kJ), convert it into Joules (J). There are 1,000 Joules in a kilojoule.
[tex]\[ \text{Heat absorbed} = 19.89 \, \text{kJ} \times 1000 = 19890 \, \text{J} \][/tex]
2. Understand the specific heat capacity of water:
The specific heat capacity of water is [tex]\(4.186 \, \text{J/g°C}\)[/tex]. This means that 4.186 joules of energy are required to raise the temperature of 1 gram of water by 1 degree Celsius.
3. Calculate the change in temperature:
Using the formula for heat transfer:
[tex]\[ Q = mc\Delta T \][/tex]
where:
- [tex]\(Q\)[/tex] is the heat absorbed,
- [tex]\(m\)[/tex] is the mass of the sample,
- [tex]\(c\)[/tex] is the specific heat capacity,
- [tex]\(\Delta T\)[/tex] is the change in temperature.
Rearrange the formula to solve for [tex]\(\Delta T\)[/tex]:
[tex]\[ \Delta T = \frac{Q}{mc} \][/tex]
Substitute the given values:
[tex]\[ \Delta T = \frac{19890 \, \text{J}}{1468.7 \, \text{g} \times 4.186 \, \text{J/g°C}} \][/tex]
[tex]\[ \Delta T \approx 3.2352099101457448 \, \text{°C} \][/tex]
4. Calculate the final temperature:
Add the change in temperature to the initial temperature to find the final temperature:
[tex]\[ \text{Initial temperature} = 7 \, \text{°C} \][/tex]
[tex]\[ \text{Final temperature} = \text{Initial temperature} + \Delta T \][/tex]
[tex]\[ \text{Final temperature} = 7 \, \text{°C} + 3.2352099101457448 \, \text{°C} \][/tex]
[tex]\[ \text{Final temperature} \approx 10.235209910145745 \, \text{°C} \][/tex]
Thus, the final temperature of the water sample will be approximately 10.24°C, when rounded to two decimal places.
1. Convert the heat absorbed into Joules:
Since the heat absorbed is given in kilojoules (kJ), convert it into Joules (J). There are 1,000 Joules in a kilojoule.
[tex]\[ \text{Heat absorbed} = 19.89 \, \text{kJ} \times 1000 = 19890 \, \text{J} \][/tex]
2. Understand the specific heat capacity of water:
The specific heat capacity of water is [tex]\(4.186 \, \text{J/g°C}\)[/tex]. This means that 4.186 joules of energy are required to raise the temperature of 1 gram of water by 1 degree Celsius.
3. Calculate the change in temperature:
Using the formula for heat transfer:
[tex]\[ Q = mc\Delta T \][/tex]
where:
- [tex]\(Q\)[/tex] is the heat absorbed,
- [tex]\(m\)[/tex] is the mass of the sample,
- [tex]\(c\)[/tex] is the specific heat capacity,
- [tex]\(\Delta T\)[/tex] is the change in temperature.
Rearrange the formula to solve for [tex]\(\Delta T\)[/tex]:
[tex]\[ \Delta T = \frac{Q}{mc} \][/tex]
Substitute the given values:
[tex]\[ \Delta T = \frac{19890 \, \text{J}}{1468.7 \, \text{g} \times 4.186 \, \text{J/g°C}} \][/tex]
[tex]\[ \Delta T \approx 3.2352099101457448 \, \text{°C} \][/tex]
4. Calculate the final temperature:
Add the change in temperature to the initial temperature to find the final temperature:
[tex]\[ \text{Initial temperature} = 7 \, \text{°C} \][/tex]
[tex]\[ \text{Final temperature} = \text{Initial temperature} + \Delta T \][/tex]
[tex]\[ \text{Final temperature} = 7 \, \text{°C} + 3.2352099101457448 \, \text{°C} \][/tex]
[tex]\[ \text{Final temperature} \approx 10.235209910145745 \, \text{°C} \][/tex]
Thus, the final temperature of the water sample will be approximately 10.24°C, when rounded to two decimal places.
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