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3 years ago

What is the difference between the specific heat of a substance and the total amount of heat of that substance ?

Physics

1 answer:

Dominik [7]3 years ago

6 0

Explanation:

Specific heat of a substance is defined as the amount of heat that is required to raise the temperature of one mole or mass of a substance by 1°C.

The total amount of heat of a substance is the amount of heat required to raise the temperature of the given mass of the substance by 1°C.

Specific heat is an intensive property and does not depend on the amount of matter that is present within a substance.

Total amount of heat is an extensive property of matter and it is predicated on the amount of matter present.

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Some say one year and a half

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A circuit contains a 1.5v battery and a bulb with a 3 ohms resistance, calculate the current of the circuit

alexandr1967 [171]

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Explanation:V=IR

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4 years ago

What does the model below represent?

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Answer:

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2 years ago

1) When making a digital animation of a person running on a sidewalk in a scene, which parameter would be an initial condition?

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3 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

3 years ago