Physics, 12.04.2020 19:16 sylvia47

A 1.00×1 0 2 g aluminum block at 100.0°C is placed in 1.00×1 0 2 g of water at 10.0°C. The final temperature of the mixture is 26.0°C. What is the specific heat of the aluminum?
CA=(-mBCB∆TB)/mA∆TA

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Answer from: 22savage2017

The answer to your question is Ca = 0.904 J/g°C

Explanation:

Data

Aluminum Water

mass 1 x10² g 1 x 10² g

Temperature 1 100°C 26°C

Temperature 2 10°C 10°C

Specific heat ? 4.182 J/g°C

Formula

Aluminum Water

-mCa(T2 - T1) = mCw(T2 - T1)

-Solve for Ca

Ca = [mCw(T2 - T1)] / -m(T2 - T1)

-Substitution

Ca = [1 x 10² x 4.182 x (26 - 10] / -1 x10²(26 - 100)

-Simplification

Ca = [418.2(16)] / -1 x 10²(-74)

Ca = 6691.2 / 7400

-Result

Ca = 0.904 J/g°C

Answer from: Quest

The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: $k= \frac{1}{2}mv^2$ where m is the mass of the arrow and v its velocity. all the other options, instead, do not increase the kinetic energy of the system: - pulling farther back on the string : this will just increase the elastic potential energy of the system- holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)- decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)

Answer from: Quest

1) estimating temperature to the nearest tenth of a degree

2) estimating the mass of the sample to the nearest tenth of a gram

4) the position of the cups of sand and water under the heat lamp

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A 1.00×1 0 2 g aluminum block at 100.0°C is placed in 1.00×1 0 2 g of water at 10.0°C. The final tem...