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Nuetrik [128]
2 years ago
7

How much heat is required to raise 100 grams of water (c= 4.18) by 5 degrees Celsius?

Physics
1 answer:
Andrei [34K]2 years ago
7 0

Answer:

Heat capacity, Q = 2090 Joules.

Explanation:

Given the following data;

Mass = 100 grams

Specific heat capacity = 4.18 J/g°C.

Temperature = 5°C

To find the quantity of heat required;

Heat capacity is given by the formula;

Q = mct

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

t represents the temperature of an object.

Substituting into the formula, we have;

Q = 100*4.18*5

Heat capacity, Q = 2090 Joules.

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Answer:

240 kPa

Explanation:

The ideal gas law states:

pV=nRT

where

p is the gas pressure

V is the gas volume

n is the number of moles

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For a fixed amount of gas, n and R are constant, so we can rewrite the equation as

\frac{pV}{T}=const.

For the gas in the problem, which undergoes a transformation, this can be rewritten as

\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}

where we have:

p_1 = 98 kPa=9.8\cdot 10^4 Pa is the initial pressure

V_1 = 750 mL=0.75 L=0.75\cdot 10^{-3} m^3 is the initial volume

T_1 =30^{\circ}C =303 K is the initial temperature

p_2 is the final pressure

V_2=250 mL=0.25 L=0.25\cdot 10^{-3} m^3 is the final volume

T_2=-25^{\circ}C=248 K is the final temperature

Solving the formula for p2, we find the final pressure of the gas:

p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}=\frac{(9.8\cdot 10^4 Pa)(0.75\cdot 10^{-3}m^3)(248 K)}{(303 K)(0.25\cdot 10^{-3} m^3)}=2.4\cdot 10^5 Pa = 240 kPa

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A 800N student does a handstand with both hands at an angle of 15 degrees from the vertical what is the force on each hand
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