questions 1/ Your team needs 250 ml of 0.085% salt solution to test your salinit

Question

questions

1/Your team needs 250 ml of 0.085% salt solution to test your salinity sensor. Identify the volumes of each of the water sources needed to create your sample. The sources are deionized water and 0.15% salt solution. (response must be in ml)

2/A tank holds 60 liters of water and the tank heater's power rating is 1300 watts (J/s). How much will the temperature of the tank change if the heater is on for 3 minutes? The Cp for water is 4.184 J/g-C.

3/You have been asked to prepare a salt solution using 18.5 grams of NaCl and 6 gallons of water, What will the weight percent salt be in the resulting solution? 1 gallon = 3.78 liters. (answer in % -

4/From data collected for RCX Display vs Actual Temperature (degrees C) a trendline is developed. Identify the sensitivity of the sensor in display unit per degrees C given the trendline shown. (DO NO INCLUDED UNITS IN THE ANSWER)

RCX Display  = 334x + 44.8, when x is actual temperature in degrees C

Thank you

Explanation / Answer

1)
total quantity of salt required = 0.085*250 = 21.25

so volume of 0.15% salt solution required = 21.25 / 0.15 = 141.667 ml

total volume = 250 ml ..

so volume of deionised water required = 250 - 141.6667 = 108.333 ml

2)
heat given by heater = power * time = 1300 * 3*60 = 234000 Joules...

for water... mass * cp * temp change

mass = 60 kg

so ... 60 * 4184 * temp change = 234000

so temp change = 0.93212 degree celcius



3)
1 gallon = 3.78 litre water = 3.78 kg
6 gallon = 6*3.78 = 22.68 kg = 22680 gram


so total mass = 22680 + 18.5 = 22698.5 gram

so weight percentage =( 18.5 / 22698.5 ) *100 = 0.08150318 %


4)
sensitivity = change in reading / change in temperature

let x1 & x2 be two temperatures..

so sensitivity = [ (334x2 + 44.8) - ( 334 x1 + 44.8 ) ] / [ x2 - x1 ] = 334* ( x2 - x1 ) / ( x2 - x1 )

so sensitivity = 334

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.