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Sagot :
### Part C - Ethanol
We start with the initial temperature of the ethanol, which is [tex]\(25.0^\circ C\)[/tex], and it absorbs 1.98 × 10³ J of heat. The final temperature recorded for ethanol is [tex]\(44.4^\circ C\)[/tex].
1. Temperature Change [tex]\( (\Delta T) \)[/tex]:
- Initial temperature [tex]\( T_i = 25.0^\circ C \)[/tex]
- Final temperature [tex]\( T_f = 44.4^\circ C \)[/tex]
- Change in temperature [tex]\( \Delta T = T_f - T_i = 44.4^\circ C - 25.0^\circ C = 19.4^\circ C \)[/tex]
2. Calculate the mass [tex]\( m \)[/tex] using the formula [tex]\( Q = mc\Delta T \)[/tex]:
- [tex]\( Q \)[/tex]: Heat absorbed = 1.98 × 10³ J
- [tex]\( c \)[/tex]: Specific heat capacity of ethanol = 2.42 J/(g·°C)
- Rearrange the formula to solve for mass [tex]\( m \)[/tex]:
[tex]\[ m = \frac{Q}{c \Delta T} = \frac{1.98 \times 10^3 \, J}{2.42 \, J/(g \cdot ° C) \times 19.4^\circ C} \][/tex]
3. Mass Calculation:
[tex]\[ m \approx 42.174 \, g \][/tex]
Therefore, the mass of the ethanol is [tex]\( \boxed{42.174} \)[/tex] grams.
### Part D - Water
Now, we consider the initial temperature of the water, which is also [tex]\(25.0^\circ C\)[/tex], and it absorbs 1.98 × 10³ J of heat. The final temperature recorded for water is [tex]\(32.4^\circ C\)[/tex].
1. Temperature Change [tex]\( (\Delta T) \)[/tex]:
- Initial temperature [tex]\( T_i = 25.0^\circ C \)[/tex]
- Final temperature [tex]\( T_f = 32.4^\circ C \)[/tex]
- Change in temperature [tex]\( \Delta T = T_f - T_i = 32.4^\circ C - 25.0^\circ C = 7.4^\circ C \)[/tex]
2. Calculate the mass [tex]\( m \)[/tex] using the formula [tex]\( Q = mc\Delta T \)[/tex]:
- [tex]\( Q \)[/tex]: Heat absorbed = 1.98 × 10³ J
- [tex]\( c \)[/tex]: Specific heat capacity of water = 4.18 J/(g·°C)
- Rearrange the formula to solve for mass [tex]\( m \)[/tex]:
[tex]\[ m = \frac{Q}{c \Delta T} = \frac{1.98 \times 10^3 \, J}{4.18 \, J/(g \cdot ° C) \times 7.4^\circ C} \][/tex]
3. Mass Calculation:
[tex]\[ m \approx 64.011 \, g \][/tex]
Therefore, the mass of the water is [tex]\( \boxed{64.011} \)[/tex] grams.
We start with the initial temperature of the ethanol, which is [tex]\(25.0^\circ C\)[/tex], and it absorbs 1.98 × 10³ J of heat. The final temperature recorded for ethanol is [tex]\(44.4^\circ C\)[/tex].
1. Temperature Change [tex]\( (\Delta T) \)[/tex]:
- Initial temperature [tex]\( T_i = 25.0^\circ C \)[/tex]
- Final temperature [tex]\( T_f = 44.4^\circ C \)[/tex]
- Change in temperature [tex]\( \Delta T = T_f - T_i = 44.4^\circ C - 25.0^\circ C = 19.4^\circ C \)[/tex]
2. Calculate the mass [tex]\( m \)[/tex] using the formula [tex]\( Q = mc\Delta T \)[/tex]:
- [tex]\( Q \)[/tex]: Heat absorbed = 1.98 × 10³ J
- [tex]\( c \)[/tex]: Specific heat capacity of ethanol = 2.42 J/(g·°C)
- Rearrange the formula to solve for mass [tex]\( m \)[/tex]:
[tex]\[ m = \frac{Q}{c \Delta T} = \frac{1.98 \times 10^3 \, J}{2.42 \, J/(g \cdot ° C) \times 19.4^\circ C} \][/tex]
3. Mass Calculation:
[tex]\[ m \approx 42.174 \, g \][/tex]
Therefore, the mass of the ethanol is [tex]\( \boxed{42.174} \)[/tex] grams.
### Part D - Water
Now, we consider the initial temperature of the water, which is also [tex]\(25.0^\circ C\)[/tex], and it absorbs 1.98 × 10³ J of heat. The final temperature recorded for water is [tex]\(32.4^\circ C\)[/tex].
1. Temperature Change [tex]\( (\Delta T) \)[/tex]:
- Initial temperature [tex]\( T_i = 25.0^\circ C \)[/tex]
- Final temperature [tex]\( T_f = 32.4^\circ C \)[/tex]
- Change in temperature [tex]\( \Delta T = T_f - T_i = 32.4^\circ C - 25.0^\circ C = 7.4^\circ C \)[/tex]
2. Calculate the mass [tex]\( m \)[/tex] using the formula [tex]\( Q = mc\Delta T \)[/tex]:
- [tex]\( Q \)[/tex]: Heat absorbed = 1.98 × 10³ J
- [tex]\( c \)[/tex]: Specific heat capacity of water = 4.18 J/(g·°C)
- Rearrange the formula to solve for mass [tex]\( m \)[/tex]:
[tex]\[ m = \frac{Q}{c \Delta T} = \frac{1.98 \times 10^3 \, J}{4.18 \, J/(g \cdot ° C) \times 7.4^\circ C} \][/tex]
3. Mass Calculation:
[tex]\[ m \approx 64.011 \, g \][/tex]
Therefore, the mass of the water is [tex]\( \boxed{64.011} \)[/tex] grams.
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