Question

value of \psi is 75 1)Estimate the rate of heat removal required to cool (500 + 10\psi) kg per hour of watermelons from 35°C to 5°C (in kW). $\dot{Q}$ = ______ kW ($\pm$ 2%) 2)Estimate the rate of heat removal required to cool (500 + 10\psi) lb$_m$ per hour of watermelons from 85°F to 35°F (in Btu/hr). Hint: see Table A-3 and A-3E $\dot{Q}$ = ______ Btu/hr ($\pm$ 2%)

          value of \psi is 75
1)Estimate the rate of heat removal required to cool (500 + 10\psi) kg per hour of
watermelons from 35°C to 5°C (in kW).
$\dot{Q}$ = ______ kW ($\pm$ 2%)
2)Estimate the rate of heat removal required to cool (500 + 10\psi) lb$_m$ per hour of
watermelons from 85°F to 35°F (in Btu/hr).
Hint: see Table A-3 and A-3E
$\dot{Q}$ = ______ Btu/hr ($\pm$ 2%)

value of ψis 75
1)Estimate the rate of heat removal required to cool (500 + 10ψ) kg per hour of
watermelons from 35°C to 5°C (in kW).
Q̇ = kW (± 2%)
2)Estimate the rate of heat removal required to cool (500 + 10ψ) lbm per hour of
watermelons from 85°F to 35°F (in Btu/hr).
Hint: see Table A-3 and A-3E
Q̇ = Btu/hr (± 2%)

Added by Ver-Nica B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Pranay Shrivastava

Step 1

18 kJ/kg°C) ΔT is the change in temperature (35°C - 5°C = 30°C) Plugging in the values, we get: Q = (500 + 10w) * 4.18 * 30 Simplifying the equation, we get: Q = 12540 + 125.4w So, the rate of heat removal required to cool (500 + 10w) kg per hour of Show more…

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The value of w is 75. 1) Estimate the rate of heat removal required to cool (500 + 10w) kg per hour of watermelons from 35Â°C to 5Â°C (in kW). Q kW (2%) 2) Estimate the rate of heat removal required to cool (500 + 10w) lbm per hour of watermelons from 85Â°F to 35Â°F (in Btu/hr). Btu/hr (2%) Hint: see Table A-3 and A-3E