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sattari [20]
3 years ago
8

Calculate the number of joules released when 72.5 grams of water at 95.0 degrees Celsius cools to a final temperature of 28.0 de

grees Celsius.
Chemistry
1 answer:
Airida [17]3 years ago
7 0
Water's specific heat capacity is 4200 J/Kg°C
95-28=67
72.5grams in kg is 0.0725kg
Energy = 67×0.0725×4200
Energy = 20,401.5 J or 20.4015 kJ
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(a) Show that the pressure exerted by a fluid P (in pascals) is given by P = hdg, where h is the column of the fluid in metres,
Komok [63]

Answer:

B) THE DEPTH OF THE LAKE IS 0.060 m

Explanation:

b) Determine the depth of the lake in metres

1. Using the general gas law, we will calculate the initial pressure of the air bubbles.

P1V1 /T1 = P2V2/T2

P1 = Unknown

T1 = 5.24 °C

T2 = 18.73 °C

P2 = 0.973 atm

V1 = V1

V2 = 6V1

P1 = P2 V2 T1 / V1 T2

P1 = 0.973 * 6V1 * 5.24 / V1 * 18.73

P1 = 5.09852 * 6 / 18.73

P1 = 30.59112 / 18.73

P1 = 1.633 atm.

2. Calculate the depth of the lake:

             Pressure = length * density * acceleration

            length = Pressure / density * acceleration

Pressure = 1.633 atm = 1.633 * 101, 325 Nm^2 = 165, 463.725 Nm^2

Density =  1.02 g/cm3 = 1.02 * 10^3 kg/m^3

Acceleration = 9.8 m/s^2

So therefore, the length in metres is:

Length = density * acceleration / pressure

Length =  1.02 *10^3 * 9.8 / 165, 463.725  

Length = 9.996 * 10^3 / 165 463.725

Length = 0.06 m

Hence, the depth of the lake is 0.06 m

6 0
3 years ago
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