Find accurate and reliable answers to your questions on IDNLearn.com. Receive prompt and accurate responses to your questions from our community of knowledgeable professionals ready to assist you at any time.
Sagot :
If the freezer operates for 6 hours each day then the power it requires for operating is (p) = 401.82W
If the freezer keeps its interior at a temperature of -6.0C in a 20.0C room, the theoretical maximum performance coefficient is (k) = 7.1
the theoretical maximum amount of ice this freezer could make in an hour, starting with water at 20.0C(m) = 24.58Kg
(a) Given energy of freezer (E) per year= 880 kW*h = 880x10^3 w*h
Time to operate = 6hrs per day
Power (P) = 880x10^3/(365x6) = 880x10^3/2190 = 401.82W
(b) Given T = -6+273 = 257K and T' = 20+273 = 293K
Coefficient(k) = T/T'-T = 257/293-257 = 7.1
(c) Q = mCwdT+mL
Q = 1kg(4190)(293-273)+1kg(3.34x10^3) = 417800J
heat removal(Qrem) = powerxk
Qrem = 401.82x7.1 =2852.922J
Mass of ice = Qrem/Q = 2852.922x3600/417800 = 24.58kg
To learn more about energy click here https://brainly.com/question/1932868
#SPJ4
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Thanks for visiting IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more helpful information.