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2 years ago

The heat capacity of 0.125Kg of water is measured to be 523j/k at a room temperature.Hence, calculate the heat capacity of water

Physics

1 answer:

Naily [24]2 years ago

5 0

Answer:

A. 4148 J/K/Kg

B. 4148 J/K/L

Explanation:

A. Heat capacity per unit mass is known as the specific heat capacity, c.

C = Heat capacity/mass(kg)

C = (523 J/K) / 0.125 Kg = 4148 J/K/Kg

B. Volume of water = mass/density

Density of water = 1 Kg/L

Volume of water = 0.125 Kg/ 1Kg/L

Volume of water = 0.125 L

Heat capacity per unit volume = (523 J/K) / 0.125 L

Heat capacity per unit volume = 4148 J/K/L

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2 years ago

A container of gas is at a pressure of 1.3x10^5 Pa and a volume of 6 m^3. How much work is done by the gas if it expands at a co

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2 years ago

1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

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$\tau=Fd$