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3 years ago

How much heat is released when 30 g of water at 96°C cools to 25°C?The specific heat of water is 1 cal/gºC.

Physics

2 answers:

Free_Kalibri [48]3 years ago

5 0

The answer is.0.213 cal and this is the correct answer for the question

Blababa [14]3 years ago

3 0

2130 cal is how much is released qhen 30 g of water at

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A 1.15-kg mass oscillates according to the equation where x is in meters and in seconds. Determine (a) the amplitude, (b) the fr

ANEK [815]

The complete question is;

A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) the amplitude, (b) the frequency, (c) the total energy, and (d) the kinetic energy and potential energy when x = 0.360 m.

Answer:

A) Amplitude; A = 0.650 m

B) Frequency; f = 1.337 Hz

C) total energy = 17.142 J

D) Kinetic energy = 11.884 J

Potential Energy = 5.258 J

Explanation:

We are given;

Mass;m = 1.15 kg

Equation; x = 0.650 cos (8.40t)

(a) The standard form of a wave function is in the form y(x,t) = Asin(kx−ωt+ϕ)

So, comparing terms in our equation in the question to this, the amplitude is;

A = 0.650 m

(b) we know that formula for frequency is;

f = ω/2π

Again, comparing terms in the standard equation and our question, we can see that ω = 8.4

Thus;

f = 8.4/(2π)

f = 1.337 Hz

E = m•ω²•A²/2

Plugging in the relevant values, we have;

E = (1.15)(8.40)²(0.650)²/2

E = 17.142 J

(d) we want to find the kinetic energy and potential energy when x = 0.360 m.

The formula for kinetic energy in this case is given by;

K = (1/2)•m•ω²•(A² - x²)

Thus;

K = (1/2) × (1.15) × (8.40)² × ((0.650)² - (0.360)²)

K = 11.884 J

Also, the formula for the potential energy in this case is given by;

U = (1/2)•m•ω²•x²

Thus;

U = (1/2) × (1.15) × (8.40)² × (0.360)²

U = 5.258 J

3 0

3 years ago

How much water

nevsk [136]

Answer:

The mass is $m = 3.75 \ kg$

Explanation:

From the question we are told that

The initial temperature is $T_1 = 15^oC$

The final temperature is $T_2 = 100 ^o C \ (boiling point )$

Generally the maximum heat produced by 1 Liter of natural gas is $9000 \ cal$

So the amount of heat produced by 100 L is

$E = 9000 * 100$

=> $E = 9000 00 \ cal$

Generally given that the efficiency is $\eta = 0.35$

Then actual heat received by the water is

$H = 0.35 * E$

=> $H = 0.35 * 9000 00$

=> $H = 315000 \ cal$

Converting to kcal

=> $H = 315000 \ cal = \frac{315000}{1000} = 315 \ kcal$

Generally the specific heat of water is

$c_w = 1 kcal/ kg \cdot ^oC$

Generally the heat received by the water is mathematically represented as

$H = m * c_w * (T_2 - T_1)$

=> $315 = m * 1 * ( 100 - 15 )$

=> $m = 3.75 \ kg$

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Which describes the relationship between potential and kinetic energy of a ball thrown up in the air as it falls back to the gro

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The answer is c, because ball is falling so its gravitationl potential energy decreases, but it kinetic energy increases. Energy is always conserved.

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Your answer is false since science is a subject not a religion

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Answer:

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