Answer:Antifreeze/coolant
Explanation: keeps your engine cool in warm weather and keeps it from freezing up in the winter. A 50-50 mix of full strength coolant and water generally protects to around -30 degrees Fahrenheit. Make sure you check with the supplier or your owner's manual for the correct formulation
Structure Of The Atom: Our current model of the atom can be broken down into three constituents parts – protons, neutron, and electrons. Each of these parts has an associated charge, with protons carrying a positive charge, electrons having a negative charge, and neutrons possessing no net charge.
Answer:
c. Detail designation
Explanation:
the question "enlarged detail to show more information", That has a detail indicator area circle. It's usually more of a square or rectangle not a circle. Circles are good for detailing landscape.
"They also can indicate when a section of the house is being enlarged." Sections details use that symbol but have a section indicator line. Also grid and column indicators use that symbol.
Answer:
Your question has some missing information below is the missing information
Given that ( specific heat of fluid A = 1 kJ/kg K and specific heat of fluid B = 4 kJ/kg k )
answer : 300 kW , 95°c
Explanation:
Given data:
Fluid A ;
Temperature of Fluid ( Th1 ) = 420° C
mass flow rate (mh) = 1 kg/s
Fluid B :
Temperature ( Tc1) = 20° C
mass flow rate ( mc ) = 1 kg/s
effectiveness of heat exchanger = 75% = 0.75
<u>Determine the heat transfer rate and exit temperature of fluid</u> <u>B</u>
Cph = 1000 J/kgk
Cpc = 4000 J/Kgk
Given that the exit temperatures of both fluids are not given we will apply the NTU will be used to determine the heat transfer rate and exit temperature of fluid B
exit temp of fluid B = 95°C
heat transfer = 300 kW
attached below is a the detailed solution
Answer:
178 kJ
Explanation:
Assuming no heat transfer out of the cooling device, and if we can neglect the energy stored in the aluminum can, the energy transferred by the canned drinks, would be equal to the change in the internal energy of the canned drinks, as follows:
ΔU = -Q = -c*m*ΔT (1)
where c= specific heat of water = 4180 J/kg*ºC
m= total mass = 6*0.355 Kg = 2.13 kg
ΔT = difference between final and initial temperatures = 20ºC
Replacing by these values in (1), we can solve for Q as follows:
Q = 4180 J/kg*ºC * 2.13 kg * -20 ºC = -178 kJ
So, the amount of heat transfer from the six canned drinks is 178 kJ.
