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n200080 [17]
3 years ago
7

A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 84 g of water (with an

initial temperature of 20 °C) in a calorimeter. The final temperature of the system is 52.1°C. The specific heat of water is 4.184 J/g*⁰C. What is the specific heat of the metal?
0.171
0.343
1.717
3.433
Physics
2 answers:
podryga [215]3 years ago
4 0
The piece of unknown metal is in thermal equilibrium with water such that Q of metal is equal to Q of the water. We write this equality as follows:

-Qm = Qw
Mass of metal (Cm)(ΔT) = Mass of water (Cw) (ΔT)

where C is the specific heat capacities of the materials.

We calculate as follows:

-(Mass of metal (Cm)(ΔT)) = Mass of water (Cw) (ΔT)
-68.6 (Cm)(52.1 - 100) = 42 (4.184) (52.1 - 20)
Cm = 1.717 -----> OPTION C
Klio2033 [76]3 years ago
3 0

Answer:

Option (d)

Explanation:

Let c be the specific heat of the metal.

By use of principle of caloriemetry,

Heat lost by the hot body = Heat gained by the cold body

Heat lost by metal = Heat gained by water

mass of metal x specific heat of metal x fall in temperature of metal = mass of water x specific heat of water x rise in temperature of water

68.6 x c x (100 - 52.1) = 84 x 4.184 x (52.1 - 20)

3285.94 x c = 11281.7376

c = 3.433 J/g C

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5 0
2 years ago
A ball of mass 4.5 kg moving with speed of 2.2 m/s in the +x-direction hits a wall and bounces back with the same speed in the x
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The change of the momentum of the ball is -19.8\, \frac{mkg}{s}

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It's important to note that momentum and velocity are vectors and direction matters, so if +x direction is the direction towards the wall and the -x direction away the wall \overrightarrow{v_{i}}=+2.2\, \frac{m}{s} and \overrightarrow{v_{f}}=-2.2\, \frac{m}{s} so (2) becomes:

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