Typically, water runs through the baseboard copper tubing and, therefore, fresh hot water is constantly running through the piping. However, consider a pipe where water was allowed to sit in the pipe. The hot water cools as it sits in the pipe. What is the temprature change, (ΔT), of the water if 193.0 g of water sat in the copper pipe from part A, releasing 2665 J of energy to the pipe? The specific heat of water is 4.184 (J/g)⋅∘C.

Answers

Answer 1
Answer:

Answer: The temperature of the water falls by 3.3°C

Explanation:

The heat change is related to the change in temperature by the equation

dH = m Cp dT

In this example, -2665 J = 193 g x 4.184 J/g°C x dT

so dT = -3.3 °C


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Answers

The answer would be B. 2.4 

You're welcome.

Answer:

Answer is B. Just took it!

Explanation:

What is a newton equal to in terms of units of mass and acceleration?

Answers

1 newton =
     the force that accelerates 1 kilogram of mass at the rate of 1meter per sec²

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If a cup of coffee is at 90°C and a person with a body temperature of 36'C touches it,how will heat flow between them

Answers

Answer:

the heat always transfers from high temperature to low temperature body without aid of any external energy to this law the heat transfers from cup of coffee to the person body until both bodies reaches to the equilibrium state    

Explanation:

The source of the sun's heat is A. nuclear fission.
B. nuclear disintegration.
C. nuclear fusion.
D. nuclear separation.

Answers

The core of the sun is so hot and there is so much pressure, nuclear fusion takes place:hydrogen is changed to helium. Nuclear fusion creates heat and photons (light). The sun's surface is about 6,000 Kelvin, which is 10,340 degrees Fahrenheit (5,726 degrees Celsius).

Answer:

C. nuclear fusion.

Explanation:

The core of the sun is so hot and there is so much pressure, nuclear fusion takes place: hydrogen is changed to helium. Nuclear fusion creates heat and photons (light). The sun's surface is about 6,000 Kelvin, which is 10,340 degrees Fahrenheit (5,726 degrees Celsius).

1 C is the total charge associated with 6.242 x 10¹⁸ electrons. How many electrons will pass through the cross section of a conductor if 50 μA of current flows for 5 s?

Answers

Answer:

Q = I t = 5.00E-8 C/s * 5.0 s = 2.5E-7 C   total charge due to 50 μC for 5 sec

6.242E18 electrons / C * 2.5E-7 C = 1.6E12 electrons

       

A current of 120 mA flows past a point in a circuit for 25 s. How much charge passes the point in this time?
How many electrons pass the point in this time?

Answers

a) I = 120 mA = 120 x 10^(-3) A = 0.12
A = 0.12 C/s ... [b/c 1 A = 1 C/s]
t = 25 s
I = Q/t
Q = It
Q = 0.12 * 25 = 3 C
so 3C of charge passes the point during the
25 s interval