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TiliK225 [7]
3 years ago
7

To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup, and measure its temperature to be 24.7 °C. You

then pour 47 mL of hot water, temperature = 46.1 °C, into the cup and measure the temperature every thirty seconds over a 10 minute period. You extrapolate this "cooling curve" back to the time of addition and find that the "final temperature" after mixing is 33.4 °C. What is the heat change of the cold water in Joules? Give your answer in standard notation. Do not input units.
Physics
1 answer:
drek231 [11]3 years ago
4 0

Answer : The heat change of the cold water in Joules is, 1.6\times 10^3J

Explanation :

First we have to calculate the mass of cold water.

As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.

Density=\frac{Mass}{Volume}

Mass=Density\times Volume=1g/mL\times 45mL=45g

Now we have to calculate the heat change of cold water.

Formula used :

Q=m\times c\times (T_2-T_1)

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water = 4.184J/g^oC

T_1 = initial temperature of cold water = 24.7^oC

T_2 = final temperature  = 33.4^oC

Now put all the given value in the above formula, we get:

Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC

Q=1638.036J=1.6\times 10^3J

Therefore, the heat change of cold water is 1.6\times 10^3J

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Now, according to the problem, as the density of the fluid does not change, we can write

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As according to the problem, A_{2} > A_{1}, so from the above formula v_{2} < v_{1}.

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

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