FSC 1st Year Physics Chapter 11 Exercise & Numericals Notes

In this post, I am sharing FSC 1st Year Physics Chapter 11 Solved Exercise & Numericals Notes PDF Download for the students of Intermediate Part 1. Heat and Thermodynamics is the 11th and last chapter from FSC First Year Physics Book. Students can download these 11th class physics chapter 11 solved Exercise & solved Numericals Notes PDF on their laptops or mobiles. This is a little file consisting of 08 pages. These Physics Notes for Class 11 are for all the boards working under Punjab Board like Gujranwala Board, Lahore Board, Faisalabad Board, Multan Board, Rawalpindi Board, Sargodha Board, DG Kahn Board, and Sahiwal Board. To download the complete FSC 1st Year Physics Notes PDF for free, click this link.

FSC Part 1 Physics Chapter 11 Heat and Thermodynamics Exercise & Numericals Notes PDF

11th Class Physics Chapter 11 Solved Numericals Notes PDF

A thermodynamic system undergoes a process in which its internal energy decreases by 300J. if at the same time, 120J of work is done on the system, find the heat lost by the system.

Data:

                Decrease in internal energy = DU = -300J
                decrease is denoted by = -ive sigh
                work done on the system = W +-120J
                Heat lost by system = Q =?
Calculation:

                As           DQ = DU +W
                putting the value we get
                DQ = -300J -120J = – 420J
                loss of heat = DQ = -420J Ans

336J of energy is required to melt 1 g  of ice at 0⁰C. what is the change in entropy of 30 g water at 0⁰C as it is changed to ice at 0⁰C by a refrigerator?

Solution:

Mass of water = m = 30g = 30 x 10^-3

Temlperature of water = T = 0⁰C = 0 +273 = 273K
heat of fusion of ice = H_f = 336 J/g
                                                336 x 1000J
change in entropy Ds =?
heat energy removed from water is

                DQ = -mH_f
                              = -30 x 10^-3 x 336 x1000
                      = -10080J
we know that
Ds = DQ/T = -10080/273 = -36.92jk^-1
Ds = – 36.92JK^-1

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