For the following circuit find the current the voltage drop

For the following circuit, find the current, the voltage drop, and power dissipation for each resistor by filling out the table. Use 2 sig fig. Found the following on the internet. I mixed 12g of coffee with 96g of hot water. The initial temperature of coffee was 77degreeF and water was 203degreeF. When I made the mixture, the new temperature wax 84degreeC. If I wanted to cool the mixture by another 20degreeC, how much heat would I have to remove? Can you please help me? Fluid properties can be found in the Appendix starting on page 328. The River Road Power Plant is a steam power plant that is designed to use natural gas to generate electrical power and provide it to the city of Vancouver. The population of Vancouver is about 150,000 residents with each resident consuming an average energy of 35 MJ per day. The natural gas boiler supplies 120 MW to the steam while thermal energy is rejected into the air (average 15degreeC). What is the minimum required temperature of the boiler to meet the power demands of the city?

Solution

Q!, Electrical engineering problem, need to submit separately!

Q(2) method of soln:

Heat gained by coffee = heat lost by hot water:

Cp water = 4.184 cal/gmdeg

96*4.184* ( temp diff) water = heat lost

Temp diff water = 203 F - 84 c . CONVERT ALL TO DEG Centigrade, = 95-84 =11 deg C

Temp diff of coffee= 84C-77F = 84C- 25C = 59 C

Calories lost by water: (Cp water 4.184)*DT = 4.184*96*11 =4418.3 calories

Calories gained by coffee: Mass of coffee X Sp heat X Temp diff = 12 * Cp coffee* 59 =708 C coffee

equating the heat lost by water to heat gained by coffee, Cp coffee is found:

Cp coffee = 6.24 cal/gm C

Now you want the final temp to be 20C less, ie 84-20 =64 deg C

The heat is removed from the final mixture of 12 g coffee and 96 g water

right now you have coffee water mixture at 84 deg C

diff in temp 20 deg C

Assuming sp heat of coffee in mixture is same:

[12g coffee*6.2405* 20] + [96g water *4.184*20] = 9531 cal to be removed from the total contents initially at 84 C

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Energy reqmt 150000*35 MJ /day = 5350000 MJ/day = 60.764 MJ/sec= 60.764 MW ( I J/sec=1W)

Boiler is supplying 120 MW, hence this is sufficient without losses.

Need to estimate efficiency of boiler taking Carnot efficiency : (T2-T1)/T2 where T2 =hot temp, T1 cold temp

( efficiency) * 120 MW= 60.76 MW ==> efficiency = .51 approx

Solve T1/T2 = .49

So if temp of boiler is 2* temp of air is enough!!