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A system absorbs 10kJ of heat at constan...

A system absorbs 10kJ of heat at constant volume and its temperature rises from `27^(0)C " to " 37^(0)C`. The ΔE of reaction is

A

100kJ

B

10kJ

C

0

D

1kJ

Text Solution

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The correct Answer is:
To solve the problem, we will use the first law of thermodynamics, which states that the change in internal energy (ΔE) of a system is equal to the heat absorbed (Q) by the system at constant volume, since no work is done (W = 0). ### Step-by-Step Solution: 1. **Identify the Given Data**: - Heat absorbed (Q) = 10 kJ - Initial temperature (T1) = 27°C - Final temperature (T2) = 37°C 2. **Understand the Condition**: - The process occurs at constant volume. This means that there is no change in volume (ΔV = 0), and therefore, no work is done by the system (W = 0). 3. **Apply the First Law of Thermodynamics**: - According to the first law of thermodynamics: \[ ΔE = Q - W \] - Since W = 0 at constant volume, we can simplify this to: \[ ΔE = Q \] 4. **Substitute the Values**: - We know that the heat absorbed (Q) is 10 kJ. Therefore: \[ ΔE = 10 \text{ kJ} \] 5. **Conclusion**: - The change in internal energy (ΔE) of the system is 10 kJ. ### Final Answer: \[ ΔE = 10 \text{ kJ} \]
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