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-Dominant- [34]
3 years ago
9

Please help! BRAINLIEST to right answer!!!

Chemistry
1 answer:
Shalnov [3]3 years ago
5 0

Answer:

B

Explanation:

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Use the Gizmo to mix 200 g of copper at 100 °C with 1,000 g of water at 20 °C. Record the data and calculated answers for copper
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Explanation:Use the Gizmo to mix 200 g of copper at 100 °C with 1,000 g of water at 20 °C. Record the data and calculated answers for copper in the 2 tables below. Accepted values for % error calculations can be found below these 2 tables.

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A 30-0 g sample of water at 280 K is mixed with 50.0 g of water at 330 K. How would you calculate the final temperature of the m
aliya0001 [1]

Answer:

311.25k

Explanation:

The question assumes heat is not lost to the surroundings, therefore

heat emitted from hotter sample ( q_{\ lost} )= heat absorbed by the less hotter  sample( q_{\ gain} )

The relationship between heat (q), mass (m) and temperature (t) is q = mc\Delta t

where c is specific heat capacity, \Delta t temperature change.

\Delta t = t_{\ final} - t_{\ initial}

equating both heat emitted and absorb

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where the values with subset 1 are the values of the hotter sample of water and the values with subset 2 are the values of the less hot sample of water.

C will cancel out since both are water and they have the same specific heat capacity.

so we have

-m_{1}(t_{\ final} - t_{\ 1initial})=m_{2}(t_{\ final} - t_{\ 2initial})

where m1 = 50g, t 1initial = 330, m2 = 30g, t2 initial = 280,t final (final temperature of the mixture) = ?

-50 * (t_{final} - 330) = 30 *  (t_{final} - 280)

-50t_{final} + 16500 = 30t_{final} - 8400

80t_{final} = 16500+8400

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