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Arceh FIlO Gqueent 401 AJ)mCi ?Wette tett> nC Fnni @ho patumgn 63 Dhoniotehn [raLC €EeeWietoMacBook Aif...

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Arceh FIlO Gqueent 401 AJ)mCi ?Wette tett> nC Fnni @ho patumgn 63 Dhoniotehn [raLC €EeeWietoMacBook Aif

Arceh FIlO Gqueent 401 AJ)mCi ? Wette tett> nC Fnni @ho patumgn 63 Dhoniotehn [ra LC € Eee Wieto MacBook Aif



Calculus 1 / AB
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An apple with an average mass of $0.18 \mathrm{kg}$ and average specific heat of $3.65 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}$ is cooled from $22^{\circ} \mathrm{C}$ to $5^{\circ} \mathrm{C} .$ The amount of heat transferred from the apple is $(a) 0.85 \mathrm{kJ}$ (b) $62.1 \mathrm{kJ}$ $(c) 17.7 \mathrm{kJ}$ $(d) 11.2 \mathrm{kJ}$ $(e) 7.1 \mathrm{kJ}$

So here in this problem were given that the average moss off the apple is 0.412 kg on its average specific heat See were given US 3.65 Kill a jewel over K G degrees Celsius on in school from temperature T even is 25 degrees Celsius. So the final temperature Tito were given us five degrees Celsius. And we have to tell them dropping change of the apple and there are five options given and we have to tell that which option is correct. So then dropping change off the apple that is Delta's is given by the formula EMC Natural log T to divide by t one so mass we have 0.412 kg times. See, we have 3.65 kill a jewel over cagey degree Celsius times Natural lock t two we have ah, five degrees Celsius. That means to 78 Calvin divide by two Anyone did you sell? Just means you to convert this thing into Kelvin. So we'll get to 98 Kelvin. So, after solving this will get the entropy change off the apple as nearly negative zero point 03042 Kill a jewel over Kelvin. So we can say that the option C is correct. So I hope you got the problem. Thank you.

Solving party of this problem. So here I will use the formula F. Any equal to G M and M A by R&A Square which is equal to 6.674. Multiplication 10 to depart -11. Multiplication, 80 kg, multiplication Euro 800.1 kg by 50 multiplication tend to depart, managed to Police squad which is required to 2.14 multiplication tend to depart minus for Newton. Now going to solve part B. So in part B I can devalue of F which can be returned F. S. A. Is equal to G M S. I am A by R S A square which is equal to 6.674 multiplication 10 to the minus 11 multiplication 1.99 Multiplication 10 to the power 30 KT multiplication 300.1 by 1.5 multiplication 10 to the power plus 11. Holy square which is equal to 5.93 multiplication 10 to the power minus four newton. As our answer for part B. Now solving party. So in part C, I can write the expression Fc, which is equal to m a g, which is equal to 0.1 kg multiplication and 9.81 m per second squared, which is equal to 0.981 newton as our answer.

With potassium Fellows I night Sweet potassium Barbados I night with potassium pharaoh Sign I'd solution. See you two platforms. See You. two platforms. See you too. F E. CN six chocolate colored PPT. Chocolate Good lad PPT. So according to the option in this problem, Option C option C Eat correct. And said for this problem with potassium Ferdowsi night solution, see you two plus form. See you to Ephesians six chocolate colored PPT. So I hope you understand the solution of this problem.

Amount off heads. Last party degree centigrade equals 4.5 jewels, but disintegrate. He's 1.51 point eight baby fell night. So we find that Thank you. Over bigly Fahrenheit is 4.5 kilo jewels over 1.8. It was 2.5. You know Jordan.

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