One kilogram of water at 0°C is heated to 100 °C. Compute its change in entropy. Given: Specific heat of water is 4.190 J/g.K. (a) 2906 J/K (b) 1400 J/K (c) 1308 J/K
Added by Enrique H.
Step 1
190 J/g.K), and ΔT is the change in temperature (100°C - 0°C = 100 K). Given: m = 1 kg c = 4.190 J/g.K ΔT = 100 K Q = 1 kg * 4.190 J/g.K * 100 K Q = 4190 J Therefore, the heat energy absorbed by the water is 4190 J. Show more…
Show all steps
Your feedback will help us improve your experience
Madhur B and 96 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Suppose 1kg of water at 100C is placed in thermal contact with 1kg of water at 0C. What is the total change in entropy? Assume that the specific heat of water is constant at 4190 J/(kg K) over this temperature range
Sri K.
Suppose 1.00 kg of water at 39.5°C is placed in contact with 1.00 kg of water at 19.5°C. (A) What is the change in energy (in joules) of the hot water due to the heat transfer when it is placed in contact with the cold water and allowed to reach equilibrium? (B) What is the change in entropy in joules per kelvin due to this heat transfer?
Madhur L.
(II) Calculate the change in entropy of 1.00 $\mathrm{kg}$ of water when it is heated from $0^{\circ} \mathrm{C}$ to $75^{\circ} \mathrm{C}$ (a) Make an estimate; (b) use the integral $\Delta S=\int d Q / T .$ (c) Does the entropy of the surroundings change? If so, by how much?
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD