Answer:
a) the COP of this air conditioner is 2.38
b) the rate of heat transfer to the outside air is 1065 kJ/min
Explanation:
Given the data in the question;
[ Outdoor ] ← Q [ W ] Q ← [ House ]
Rate of heat removed from the house; Q = 750 kJ/min = ( 750 kJ/min × ( 1 kW / 60 kJ/min ) ) = 12.5 kW
Net-work input; W = 5.25 kW
a) The coefficient of performance of the air conditioner; COP.
COP = Q / W
we substitute
COP = 12.5 kW / 5.25 kW
COP = 2.38
Therefore, the COP of this air conditioner is 2.38
b) the rate of heat transfer to the outside air.
Q = Q + W
we substitute
Q = 12.5 kW + 5.25 kW
Q = 17.75 kW
Q = ( 17.75 × 60 ) kJ/min
Q = 1065 kJ/min
Therefore, the rate of heat transfer to the outside air is 1065 kJ/min
Answer:
If your spent lead-acid batteries will be reclaimed through regeneration, you are exempt from all of the hazardous waste regulations except for Part 261 and §262.11 (both of which are waste identification requirements). ... The Universal Waste rules in Part 273 apply to all types of batteries that would be hazardous waste
Explanation:
Answer:
true
Explanation:
if it is not true it is false
Answer:
1.636
Explanation:
Slope is given by change in y coordinates against change in x coordinates. Here, change in y will be 973.738-660.974= 312.764 and change in x will be 437.279-246.060=191.219
The slope, m will be 312.764/191.219=1.6356324423828
Rounding off, the slope is 1.636 to four significant figures