Answer :
To find the amount of absorbed heat, we need to follow these steps:
1. Convert the final temperature to Kelvin:
- The initial temperature is given in Kelvin: [tex]\( T_{\text{initial}} = 300 \text{ K} \)[/tex].
- The final temperature is given in degrees Celsius: [tex]\( T_{\text{final}} = 300^{\circ} \text{C} \)[/tex].
- Convert the final temperature to Kelvin: [tex]\( T_{\text{final}} = 300 + 273.15 = 573.15 \text{ K} \)[/tex].
2. Calculate the temperature change ([tex]\(\Delta T\)[/tex]):
- [tex]\(\Delta T = T_{\text{final}} - T_{\text{initial}} \)[/tex]
- [tex]\(\Delta T = 573.15 \text{ K} - 300 \text{ K} = 273.15 \text{ K} \)[/tex]
3. Calculate the heat absorbed (Q) using the formula [tex]\( Q = m \cdot c \cdot \Delta T \)[/tex]:
- Mass ([tex]\( m \)[/tex]): [tex]\( 25 \text{ g} \)[/tex]
- Specific heat ([tex]\( c \)[/tex]): [tex]\( 0.45 \text{ J/(g·}^\circ \text{C)} \)[/tex]
- Temperature change ([tex]\(\Delta T\)[/tex]): [tex]\( 273.15 \text{ K} \)[/tex] (note that [tex]\(\Delta T\)[/tex] in Kelvin is equivalent to [tex]\(\Delta T\)[/tex] in °C for the purpose of this calculation)
- [tex]\( Q = 25 \text{ g} \times 0.45 \text{ J/(g·}^\circ \text{C)} \times 273.15 \text{ K} \)[/tex]
- [tex]\( Q = 25 \times 0.45 \times 273.15 \)[/tex]
- [tex]\( Q = 3072.9375 \text{ J} \)[/tex]
4. Convert the absorbed heat from Joules to Kilojoules:
- [tex]\( 1 \text{ KJ} = 1000 \text{ J} \)[/tex]
- [tex]\( Q_{\text{KJ}} = \frac{Q_{\text{J}}}{1000} \)[/tex]
- [tex]\( Q_{\text{KJ}} = \frac{3072.9375 \text{ J}}{1000} \)[/tex]
- [tex]\( Q_{\text{KJ}} = 3.0729375 \text{ KJ} \)[/tex]
So, the amount of absorbed heat is [tex]\( 3.0729375 \)[/tex] KJ.
1. Convert the final temperature to Kelvin:
- The initial temperature is given in Kelvin: [tex]\( T_{\text{initial}} = 300 \text{ K} \)[/tex].
- The final temperature is given in degrees Celsius: [tex]\( T_{\text{final}} = 300^{\circ} \text{C} \)[/tex].
- Convert the final temperature to Kelvin: [tex]\( T_{\text{final}} = 300 + 273.15 = 573.15 \text{ K} \)[/tex].
2. Calculate the temperature change ([tex]\(\Delta T\)[/tex]):
- [tex]\(\Delta T = T_{\text{final}} - T_{\text{initial}} \)[/tex]
- [tex]\(\Delta T = 573.15 \text{ K} - 300 \text{ K} = 273.15 \text{ K} \)[/tex]
3. Calculate the heat absorbed (Q) using the formula [tex]\( Q = m \cdot c \cdot \Delta T \)[/tex]:
- Mass ([tex]\( m \)[/tex]): [tex]\( 25 \text{ g} \)[/tex]
- Specific heat ([tex]\( c \)[/tex]): [tex]\( 0.45 \text{ J/(g·}^\circ \text{C)} \)[/tex]
- Temperature change ([tex]\(\Delta T\)[/tex]): [tex]\( 273.15 \text{ K} \)[/tex] (note that [tex]\(\Delta T\)[/tex] in Kelvin is equivalent to [tex]\(\Delta T\)[/tex] in °C for the purpose of this calculation)
- [tex]\( Q = 25 \text{ g} \times 0.45 \text{ J/(g·}^\circ \text{C)} \times 273.15 \text{ K} \)[/tex]
- [tex]\( Q = 25 \times 0.45 \times 273.15 \)[/tex]
- [tex]\( Q = 3072.9375 \text{ J} \)[/tex]
4. Convert the absorbed heat from Joules to Kilojoules:
- [tex]\( 1 \text{ KJ} = 1000 \text{ J} \)[/tex]
- [tex]\( Q_{\text{KJ}} = \frac{Q_{\text{J}}}{1000} \)[/tex]
- [tex]\( Q_{\text{KJ}} = \frac{3072.9375 \text{ J}}{1000} \)[/tex]
- [tex]\( Q_{\text{KJ}} = 3.0729375 \text{ KJ} \)[/tex]
So, the amount of absorbed heat is [tex]\( 3.0729375 \)[/tex] KJ.
