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1 year ago

How much heat is released when 432 g of water cools down from 71'c to 18'c?

1 answer:

7 0

The heat released by the water when it cools down by a temperature difference $\Delta T$ is
$Q=mC_s \Delta T$
where
m=432 g is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of water
$\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ}$ is the decrease of temperature of the water

Plugging the numbers into the equation, we find
$Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J$
and this is the amount of heat released by the water.

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Answer:

Explanation:

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1 year ago

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Answer:

When she adds more washers to the meter, the magnitude of force that is shown on the force meter increases.

Explanation:

The force that the washers exert on the force meter is actually the weight of the washers. Weight is actually a force with gravitation acceleration.

F = W = mg

Where g is gravitational acceleration and its value is 9.81 m/s² and m is the mass of any object. As she adds more washers to the meter so the total mass of the washers increases. As the mass of the washers increases, magnitude of the force (Weight) shown on the force meter increases.

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1 year ago

Calculate the energy in the form of heat (in kJ) required to change 75.0 g of liquid water at 27.0 °C to ice at –20.0 °C. Assume

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Explanation:

The given data is as follows.

mass, m = 75 g

$T_{1} = 0^{o}C$

$T_{2} = 27^{o}C$

Specific heat of water = 4.18

First, we will calculate the heat required for water is as follows.

q = $m C \times (T_{1} - T_{2})$

= $75 g \times 4.18 J/g^{o}C \times (0 - 27)^{o}C$

= 8464.5 J/mol

= 8.46 kJ ......... (1)

Also, it is given that $T_{3} = -20^{o}C$ = (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.

Now, we will calculate the heat of fusion as follows.

q = $mC \times (T_{3} - T_{1})$

= $0.075 kg \times 2.108 kJ/kg K \times (-293 - 0) K$

= -46.32 kJ ......... (2)

Now, adding both equations (1) and (2) as follows.

8.46 kJ - 46.32 kJ

= -37.86 kJ

Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at $27.0^{o}C$ to ice at $-20.0^{o}C$ is -37.86 kJ.

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1 year ago

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