Block Volume: LWH Cube: Cylinder. Ah Sphere: (43hr m-pV (Water: 1.0 g/cm or 1000

Question

Block Volume: LWH Cube: Cylinder. Ah Sphere: (43hr m-pV (Water: 1.0 g/cm or 1000 kg/m) P.-101.000 Pa 1 Pa= 1 N/m2 P-14.7 Ibs/in I "bar" () = 100,000 Pa 1000 "milli-bars"(mb)-"ba Innb 100 Pa Gauge Pressure = Total Pressure-P Hydraulic Lift: F2-(A2/Ai) F B- weight of fluid displaced w=mg F=ma Tc=T-273 TF (9/5) Tc+32 1 calorie (cal 4.186 joules (J) I Calorie (Cal) 1000 cal Q-mcAT Thermal Properties of H2O: cm 1.0 calg-Co (water) c 0.5 cal/g-Co (ice or steam) LF = 80 cal/g Lv-540 calg Mixtures: Q_0 H (M/Mmax) 100% Dew Point-Temperature at which water vapor in air starts to condense R-oeAT4 where -5.67 x 10-8 w/m2-K4

Explanation / Answer

Difference between the average temperature of the body between the levels and outside temp.k is a constant

dQ / dt = k ( T2-T1 )

body between the levels and outside temp.k is a constant.

In this case, the final temperature, T2=(800+700)/2 = 750 K

and initial temperature T1=600 K

Difference between temperaures, T2 - T1 = 750 K - 600 K = 150 K

Now use the equation,   dQ / dt = k ( T2-T1 )

0.20 = k ( 150 K )

Rearrange for k

k = 2 / 1500

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In the case of when it goes all the way from 700k to 600k, it is going to drop till outside temp

Fianl temperature, T2 = (700 K + 600 K ) / 2

Difference in temperature, T2 - T1 = 50 K

Now,use the equation dQ / dt = k ( T2-T1 )

dQ / dt = k ( 50 K )

Substitute k = 2 / 1500

dQ / dt = ( 2/1500 ) ( 50 K )

dQ / dt = 0.067 J/min

Thus, none of the given option is correct.

Thus, the correct option is ( E )

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