Print Calculator Periodic Table Question 7 of 17 Map Presented by Physics openst

Question

Print Calculator Periodic Table Question 7 of 17 Map Presented by Physics openstax sapling learning What is the best possible coefficient of performance for a refrigerator that cools an environment at 13.0 °C and exhausts heat to another environment at 45.0 °C? Number How much work in joules must be done for a heat transfer of 3.925 x 106 J from the cold environment? Number What is the cost of doing this, if the work costs 10.5 cents per 3.60 x 106 J (a kilowatt-hour)? Number cents How many joules of heat transfer occurs into the warm environment? Number

Explanation / Answer

(a)Given that

T1 = -13 + 273 = 260 K

T2 = 45+ 273 = 318 K

Best COP(coefficient of performance) of refergeriator is given as:

COP = T1/ T1-T2

COP = 260/ (318-260) = 4.48

Hence, coefficient of performance = COP = 4.48.

(b) Given that,

Q(supplied) = 3.925 x 106 J

We know that,

COP = Q(supplied)/ (W) Workdone => Workdone = Q(supplied) / COP

W = 3.925 x 106/4.48 = 8.76 x 105 J

Hence, Workdone = W = 8.76 x 105 J

(c) Given that,

10.5 cents Per 3.6 x 106 J

So, 8.76 x 105 Joule will costs = (10.5/ 3.60 x 106 ) x 8.76 x 105 = 2.56 cents

Hence, cost = 2.56 cents.

(d) heat transferred to the environment can be calculated as follows:

COP = Q(transfer)/ WD

4.48 = Q(transfer)/ 8.76 x 105

Q = 8.76 x 4.48 x 105 = 39.24 x 105 Joules

Hence, Q(transfer) = 39.24 x 105 Joules

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.