STANDARD Physics First Law of Thermodynamics MCQ Question
1 kg of water is heated from 40°C to 70°C. If its volume remains constant, then the change in internal energy is (specific heat of water = 4148 J kg⁻¹ K⁻¹)
2.44 × 10⁵ J
1.62 × 10⁵ J
1.24 × 10⁵ J
2.62 × 10⁵ J
Correct Answer
Detailed Explanation
Since volume of water remains constant, then work done ΔW = PdV = 0. According to first law of thermodynamics dQ = dU + dW, dU = dQ = msΔT = 1 × 4148 × (70 − 40) = 4148 × 30 = 124440 J = 1.244 × 10⁵ J.
Found an issue with this question?
Related Questions
More from Thermodynamics
A Carnot engine has an efficiency of 50% when its source is at a temperature 327° C. The temperature of the sink is :
A system goes from A to B by two different paths in the P-V diagram as shown in figure. Heat given to the system in path 1 is 1100 J, the work done by...
Among the following, the set of parameters that represents path function is: (A) q + w (B) q (C) w (D) H − TS