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zhannawk [14.2K]
3 years ago
10

An experiment was devised to investigate the effects on the temperature of a cup of coffee when cream is added at different time

intervals. Hot and cold water were used instead of coffee and cream. First, a container of water heated to 80°C was allowed to sit for 15 minutes. Its temperature was measured at regular intervals. Next, similar temperature measurements were taken for 15 minutes on a new container of heated water, but this time some cold water was added after the 2-minute reading. Finally, the procedure was repeated except that for this last test, the cold water was added after the 12-minute reading. The resulting recorded temperatures were plotted on the graph shown here.
1. What question was asked in the experiment?


2. What was the hypothesis, expressed as an if-then statement?


3. What were the independent and dependent variables?


4. What was the control in the experiment?


5. There is one thing that is wrong about this graph. What is it?
Physics
1 answer:
svet-max [94.6K]3 years ago
7 0


Hi Babylilith32,


Your Question:


An experiment was devised to investigate the effects on the temperature of a cup of coffee when cream is added at different time intervals. Hot and cold water were used instead of coffee and cream. First, a container of water heated to 80°C was allowed to sit for 15 minutes. Its temperature was measured at regular intervals. Next, similar temperature measurements were taken for 15 minutes on a new container of heated water, but this time some cold water was added after the 2-minute reading. Finally, the procedure was repeated except that for this last test, the cold water was added after the 12-minute reading. The resulting recorded temperatures were plotted on the graph shown here.  


Answers:


1. If you have cool water then you have to add it to the hot water after a few minutes.


2. The independant variable is the cold water


3. The dependant variable is the temperature


4. The control is the hot water without the cold water.


5. The 2 min cold water drops way to much.


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\begin{aligned} \text{Average speed} &= \frac{\text{Total Distance}}{\text{Total Time}}\\ &= 50 \left/\frac{35}{48}\right.\\ &= 50 \cdot \frac{48}{35} \\&= \frac{480}{7}\approx 68.6\; \rm mph \end{aligned}.

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Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water
s2008m [1.1K]

Complete Question

The complete question is shown on the first and second uploaded image

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a

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is \frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5

b

The number of moles of gas that must be released is  n= 0.3538\ mols

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

 This means that the pressure at the depth of the surface would be

                P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm

                      = 2.5 atm

The ideal gas equation is mathematically represented as

                PV = nRT

Where P is pressure at the surface

           V is the volume

            R is the gas constant  = 8.314 J/mol. K

making n the subject we have

        n = \frac{PV}{RT}

 Considering at the surface of the water the number of moles at the surface would be

               n_s = \frac{P_sV}{RT}

Substituting 1 atm = 101325 N/m^2 for P_s ,6L = 6*10^{-3}m^3 for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

              n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}

                   = 0.2359 mol  

To obtain the number of moles at the depth of the water we use

                n_d  = \frac{P_d V}{RT}

Where P_d \ and \ n_d \ are pressure and no of moles at the depth of the water

        Substituting values we have

              n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}

                  = 0.5897 mol

Now to obtain the number of moles released we have

             n =  n_d - n_s

               = 0.5897mol  - 0.2359mol

              =0.3538 \ mol

     The molar concentration at the surface  of water is

                [\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}

                                =39.31mol/m^3

    The molar concentration at the depth  of water is

           [\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}

                      = 98.28 mol/m^3

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

         \frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5

                   

                     

                     

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