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Goryan [66]
3 years ago
7

A 200 g piece of iron is heated to 100°C. It is then dropped into water to bring its temperature down to 22°C. What is the amoun

t of heat transferred to water? (ciron = 0.444 J/g°C) A) 1.9 kJ B) 6.9 kJ C) 8.9 kJ Eliminate D) 10.9 kJ
Physics
2 answers:
Amiraneli [1.4K]3 years ago
8 0
The piece of iron is brought from a temperature of 100C to a temperature of 22C. In this process, the heat released by this piece of iron is
Q=m C_s \Delta T
where m=200 g is the mass of the piece of iron, C_s = 0.444 J/(g C) is the iron's specific heat, and the temperature variation is \Delta T = 22^{\circ}-100^{\circ}=-78^{\circ}C. Using these values, the amount of heat released by the iron is
Q=(200 g)(0.441 J/(gC))(-78C)=-6926J=-6.9 kJ
With negative sign because it is amount of heat released. This heat is then trasferred to the water, so the amount of heat absorbed by the water is Q=6.9 kJ.
ASHA 777 [7]3 years ago
5 0

The answer is B) 6.9 kJ.

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At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coeffi
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The given data is incomplete. The complete question is as follows.

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36.  Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)

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Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and \mu be the friction coefficient and m be the mass of car.

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                v = 0,     s = 84 m,     \mu = 0.36

According to Newton's law of second motion the expression for acceleration is as follows.

                      F = ma

                 -\mu N = ma

                 -\mu mg = ma

                      a = -\mu g

Also,    

               v^{2} = u^{2} + 2as

              (0)^{2} = u^{2} + 2(-\mu g)s

                  u^{2} = 2(\mu g)s

                            = \sqrt{2(0.36)(9.81 m/s^{2})(84 m)}

                            = 24.36 m/s

Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.

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