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olga2289 [7]
3 years ago
9

Suppose a 10 mL beaker and a 100 mL beaker are both filled with water, and that the water in both beakers is the same temperatur

e. Which statement below about the water in the two beakers is true?
a. the two beakers of water contain equal amounts of heat
b. the larger beaker of water contains more heat than the smaller beaker
c. the smaller beaker of water contains more heat than the larger beaker
d. it is not possible to determine which beaker of water contains the most heat because heat cannot be measured
Physics
2 answers:
Doss [256]3 years ago
7 0

The correct choice is

b. the larger beaker of water contains more heat than the smaller beaker

Reason :

Heat is given as

Q = m c T

where Q = heat contained

m = mass of water

c = specific heat of water

T = change in temperature.

since the specific heat and temperature is same for both the beakers, we need to look at which beaker contains more amount of water. the larger beaker has more mass of water, hence heat contained in the larger beaker is greater.

Sergeeva-Olga [200]3 years ago
6 0
The larger beaker of water contains more heat than the smaller beaker.
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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

Answer:

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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