Question:
Combustion of 1 mole graphite releases \(2.48 \times 10^2\) kJ of energy. What will be the temperature of a bomb calorimeter if 1 g of graphite is burnt at 298 K, given heat capacity = 10.35 kJ/K?
Combustion of 1 mole graphite releases \(2.48 \times 10^2\) kJ of energy. What will be the temperature of a bomb calorimeter if 1 g of graphite is burnt at 298 K, given heat capacity = 10.35 kJ/K?
Show Hint
In bomb calorimetry, use \(q = C\Delta T\) and always convert moles before applying enthalpy values.
Updated On: Jun 19, 2026
- 298 K
- 296 K
- 300 K
- 299 K
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Understanding energy released per mole.
Given combustion of 1 mole graphite releases: \[ q = 2.48 \times 10^2 = 248 \text{ kJ} \]
Step 2: Calculating moles of graphite burnt.
Mass of graphite = 1 g, molar mass of carbon = 12 g/mol: \[ n = \frac{1}{12} = 0.0833 \text{ mol} \]
Step 3: Calculating heat released.
Heat released for 0.0833 mol: \[ q = 248 \times 0.0833 \approx 20.67 \text{ kJ} \]
Step 4: Temperature rise in calorimeter.
Using: \[ q = C \Delta T \] \[ \Delta T = \frac{20.67}{10.35} \approx 2 \text{ K} \]
Step 5: Final temperature calculation.
Initial temperature = 298 K: \[ T_{final} = 298 + 2 = 300 \text{ K} \]
Final Answer: \[ \boxed{300 \text{ K}} \]
Given combustion of 1 mole graphite releases: \[ q = 2.48 \times 10^2 = 248 \text{ kJ} \]
Step 2: Calculating moles of graphite burnt.
Mass of graphite = 1 g, molar mass of carbon = 12 g/mol: \[ n = \frac{1}{12} = 0.0833 \text{ mol} \]
Step 3: Calculating heat released.
Heat released for 0.0833 mol: \[ q = 248 \times 0.0833 \approx 20.67 \text{ kJ} \]
Step 4: Temperature rise in calorimeter.
Using: \[ q = C \Delta T \] \[ \Delta T = \frac{20.67}{10.35} \approx 2 \text{ K} \]
Step 5: Final temperature calculation.
Initial temperature = 298 K: \[ T_{final} = 298 + 2 = 300 \text{ K} \]
Final Answer: \[ \boxed{300 \text{ K}} \]
Was this answer helpful?
0
0
Top AP EAPCET Chemistry Questions
- The four quantum numbers of the valence electron of potassium are :
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
- In which of the following the election gain enthalpy of elements is correctly arranged?
- The radii of $ \text{F,}{{\text{F}}^{-}}\text{O} $ and $ {{\text{O}}^{2-}} $ are in the order
- Which of the following has the highest first ionization potential?
View More Questions
Top AP EAPCET Work done in Thermodynamics Questions
An ideal gas undergoes an isothermal process at 300 K. If the initial volume is 2 m\(^3\) and it doubles to 4 m\(^3\), what is the work done by the gas? (Take \(R = 8.31 \, \text{J/(mol.K)}\), and assume 1 mole of gas.)
- For a spontaneous process at constant temperature and pressure, the change in Gibbs free energy ($\Delta G$) is
- At 27^C, 1.6 g of O$_2$ gas at 5 atm expands isothermally against a constant external pressure of 1 atm. The work done (in J) is (1 L-atm = 100 J):
- Cp of a gas is 20 J mol\(^{-1}\) K\(^{-1}\). If one mole of the gas is cooled from 298 K to 258 K at constant pressure, calculate the internal energy change of the system. (R = 8.3 J mol\(^{-1}\) K\(^{-1}\))
- Calculate \(\Delta_r H^0\) for the reaction \(A^+ (aq) + B^- (aq) \rightarrow AB(s)\) at 25°C. Given:
\[ \Delta_f H^0 (A^+, aq) = 200~\text{kJ mol}^{-1}, \quad \Delta_f H^0 (B^-, aq) = -300~\text{kJ mol}^{-1}, \] \[ \Delta_f H^0 (A^+, g) = -150~\text{kJ mol}^{-1}, \quad \Delta_f H^0 (AB, s) = -250~\text{kJ mol}^{-1}, \quad \Delta_f H^0 (AB, g) = +130~\text{kJ mol}^{-1} \]
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions