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just olya [345]
3 years ago
15

How much heat is required to warm 122 g of water by 23.0 c?

Physics
2 answers:
Pachacha [2.7K]3 years ago
6 0

Answer: 11745.92 j

Explanation: using heat = mc∆

Where k is mass

C is heat capacity of water = 4.186 j/gc

∆ is temperature

H = 122 × 4.186 × 23

= 11745.92 j

GaryK [48]3 years ago
4 0
<span>122 g * 4,186 (j/g*°c) * 23°c = 11745.916 j </span>
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A box full of charged plastic balls sits on a table. The electric force exerted on a ball near one upper corner of the box has c
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We have that the values for F north, F east, F up are

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From the Question we are told that

electric force F_1 = 1.2 x 10^{-3} N(N)

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Generally the equation for the F north  is mathematically given as

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For F East

F_E=\frac{F_2}{q_1}\\\\F_E=\frac{5.7 x 10^-4 }{110}

F_E=5.18181818*10^{-6}

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For more information on this visit

brainly.com/question/21811998

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Two moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2250 J of heat is added to the gas, and 870 J of work i
Lapatulllka [165]

Answer: The final temperature is 470K

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ΔU =3/2nR(T2-T1)

ΔU=3/2 × 2 ×8.314 (T2 - 345)

ΔU=24.942(T2-345)

Therefore Q = 24.942(T2-345)+ (-870)

2250=24.942(T2-345)+ (-870)

125.09=(T2-345)

T2 =470K

Therfore the final temperature is 470K

5 0
2 years ago
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Answer:

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