Could you please take a look at this? I'm having trouble with #2 and #3. Thank you!
#1 - Initial temperature of Iron shot (Fe) metal: 100°C
Final temperature of Iron shot (Fe) metal: 28°C
Initial temperature of water from calorimeter: 20°C
Final water temperature from calorimeter: 28°C
#2 - Estimate of specific heat for metal:
(use 1 cal/gm-°C for water)
#3 - Estimate the atomic weight of Iron shot (Fe) metal:
Step 1: Obtain metal sample in test tube. First add a test tube to the Lab. Add 100 gm of Iron Shot (Fe). Add a thermometer to the test tube so that you can record the initial temperature of the Fe.
Step 2: Prepare hot water bath and add test tube with metal. Obtain a 250 ml beaker. Add 150ml of water at room temperature to the beaker. Place test tube within beaker. Now heat the combined beaker test tube arrangement by heating with a bunsen burner until the Fe shot reaches the boiling point of water 100 °C.
Step 3: Place heated metal in calorimeter with water at room temp. Add Calorimeter to Lab. Add 100 ml of water at room temp 20 °C to calorimeter. Remove test tube from beaker. Pour the heated Iron into the calorimeter and close the calorimeter. Record the final temperature of the water.© BrainMass Inc. brainmass.com October 25, 2018, 5:17 am ad1c9bdddf
So for #2 we want to find the specific heat, which as you recall was the "c" in q=mcdT.
We know the m of Fe is 100g. The initial T is 100 degrees, and the final T is 28 degrees.
The initial temp of the water was 20 degrees, and the final was 28. 150 mL of water is 150g ...
How do we find the molecular weight of a substance and its specific heat if we know some data about its enthalpy? In this solution we explore, step-by-step, the math involved in solving these types of problems!
3) SOLVE THIS PROBLEM BELOW:
A hothouse needs to be kept at a constant temperature of 30°C drybulb and 60% relative humidity (state K) throughout the year. The air flowrate out of the hothouse is 5 m 3 /s .
40% of this air must be replaced by outside air. In winter the air is at 8°C and 90% relative humidity (state B). There is also a heat loss of 37 kW through the "walls" of the glasshouse and the plants add moisture through evapotranspiration; this is equivalent to a latent load of 25,7 kW.
The return air ducts are not well insulated and the air temperature changes by 1,5°C (to state R) before it is mixed with the outside air.
In the plantroom (machine, not vegetation) this mixture of air (state M) is first heated (to state H) and then adiabatically humidified (to state A) before entering the supply air ducts; these ducts are also not well insulated and the air temperature changes by 2°C before reaching the hothouse (state S).
Using calculations and a psychrometric chart where necessary ...
i. Calculate the mass flow rate of the supply air to the hothouse.
ii. Determine the supply air state S at which air must be supplied. Give this
In terms of the dry-and wetbulb temperatures.
iii. Determine the state of the mixture of outside and return air before it is
treated in the plantroom. Give the state in terms of the drybulb and wetbulb temperatures. This is state M.
iv. Determine the state of the air after the adiabatic humidifier. This is state A.
Note that the air is not saturated but that sufficient water is available to
humidify the air to the desired state.
v. Determine the temperature to which the air must be heated before being adiabatically humidified. This is state H.
Show all processes on the psychrometric chart provided - the atmospheric pressure is 101,325 kPa.
CALCULATIONS IN PSYCHROMETRICS USING APPROXIMATE FORMULAE
Often psychrometric processes involving the psychrometric chart , mass flow rates and heat transfers are difficult since not all the information is available ; for instance , unless one knows where the state is one does not know what the specific volume is and since volume and massflow rate are linked to each other by the specific volume , solving the problem often leads to trial and error solutions . In order to overcome this difficulty one may use the following approximations and then redo the problem more accurately if necessary using the values obtained by the approximate forumula .
1 ) In problems concerning the mixing of flowrates the Conservation of mass law demands that mass be conserved,,,there is no " conservation of volume law "! However in most ordinary low pressure systems the density or specific volume of the air does not change that much and volumes may be added to give total volume and in mixing processes one may use
ma / mb ~ V a / Vb
also ma + mb ~ Va + Vb
2 ) Concerning changes in enthalpy as a result of the addition of heat energy
The addition or loss of SENSIBLE heat causes the dry bulb temperature to rise or fall . The following equation is applicable
mda h1 + Q sensible = mda h 2
now one may write h 2- h1 = cp atmospheric. air (t2 - t1 )
where cp atmospheric air = cp dry air + w ( cp water vapour )
and the mass flow rate may be written as
m = V / v
Most Air Conditioning applications occur in the region where the value of the specific volume can be viewed as an average of 0,83 m3/kg and where the specific humidity is 0,0073 kg/kgda. ...these values are at 17 °C dry bulb temperature and 60 % relative humidity - usually the ball park figure for supply air for most air conditioning situations .
Using these values the following equations are useful viz
m = V / 0,83
and cp atmospheric air = cp dry air + w ( cp water vapour ) = 1.007 + 0,0073(1,89)
= 1,0208 kJ/ kg kW
Putting all this together we find
QS = V/0,83*cp atmospheric air (t2 - t1 )
= V*1,2165*1,0208 (t2 - t1 ) = 1,23 V(t2 - t1 )
Q sensible = 1,23 V (t2 - t1 )dry bulb temperature
a very useful equation !!
Please note that the change in DRY BULB TEMPERATURE is only due to the SENSIBLE heat added or removed .
PSYCHROMETRICS PART TWO
DALTONS LAW OF PARTIAL PRESSURES - an important fundamental concept
Both the collection of dry air molecules ( ie. O2 , N2 , CO2 etc. ) and the water vapour are viewed as ideal gases , We may consider the dry air as a single gas called dry air and which has a gas constant of value Ra = 0,2871 kJ / kg K.
The water vapour can also be viewed as an ideal gas without too much error and it has a gas constant Rv = 0,4615 kJ / kg K . (The textbook uses subscript s for vapour and other books use subscript v so I shall use v just to get you used to either symbol. )
Consider a sample of atmospheric air which consists of 1 kg of dry air molecules and w kg of water vapour . ( Whatever psychrometric process this sample undergoes the constant is always the mass of dry air molecules that are conserved ; the water vapour is a " passenger " which may get on or off anytime ie water vapour may condense out of the atmospheric air when the air is cooled to below its dewpoint temperature or the specific humidity may increase when adiabatic saturation takes place or when a sample of air is humidified and heated by adding steam to the air .)
For each ideal gas one may write
p V = m R T
p [ kPa ] V [ m3 ] R [ kJ / kg K ] T [ K ] m [ kg. ]
a refers to the dry air
v refers to the water vapour ........ so one may write
pa Va = 1 R aTa
pv Vv = w R vTv
For the mixture as a whole one may write
patmosphere Vmixture = (1 + w ) R mixtureT mixture
Now since atmospheric air may be viewed as a mixture of two ideal gases Daltons Law of Partial Pressures may be applied . ie.
Separate ( in your mind ) the atmospheric air into the two ideal gases and place each one into a cylinder which has a massless piston on top .Now place a weight on top of each piston so that the volume occupied by each gas is the same as the volume of the original mixture and at the same temperature as the original mixture . Now the weight on top of each piston will exert a pressure on the gas which is exactly equal to the partial pressure of the gas as it exists in the mixture
For each cylinder one may write
pav = RaT
pv v = w Rv T
where v is the volume occupied by 1 kg of dry air molecules plus w kg water vapour as it exists in the atmosphere- this is known as the specific volume of atmospheric air per kg of dry air
And Daltons Law states that pa + p v= p atmosphere
Also to be noted is that the "specific volume" of the mixture is not a true specific volume since the mass of the mixture is not 1 kg , yet it is the true specific volume of the dry air at the partial pressure of the dry air .View Full Posting Details