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musickatia [10]
2 years ago
6

A 0.350 kg block at -27.5°C is added to 0.217 kg of water at 25.0°C. They come to equilibrium at 16.4°C. What is the specific he

at of the block?

Physics
1 answer:
igor_vitrenko [27]2 years ago
5 0

Answer:

508 J/kg/C

Explanation:

Energy Lost by water = Energy gained by block

mcT = <em>m</em><em>c</em><em>T</em><em> </em><em> </em>[bolded is for water, <em>i</em><em>t</em><em>a</em><em>l</em><em>i</em><em>c</em><em>i</em><em>s</em><em>e</em><em>d</em><em> </em>is for block]

(0.217)(4186)(25 - 16.4) = <em>(</em><em>0</em><em>.</em><em>3</em><em>5</em><em>0</em><em>)</em><em>(</em><em>c</em><em>)</em><em>(</em><em>1</em><em>6</em><em>.</em><em>4</em><em> </em><em>+</em><em> </em><em>2</em><em>7</em><em>.</em><em>5</em><em>)</em>

<em>1</em><em>5</em><em>.</em><em>3</em><em>6</em><em>5</em><em>c</em><em> </em>= 7811.9132

c = <u>5</u><u>0</u><u>8</u><u> </u><u>J</u><u>/</u><u>k</u><u>g</u><u>/</u><u>C</u><u> </u><u>(</u><u>3</u><u> </u><u>s</u><u>f</u><u>)</u>

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mass \quad of\quad moon = m_{1}\\\\mass\quad of \quad comet = m_{2} = 0.5 m_{1}\\\\speed\quad of\quad moon\quad before\quad collision = v_{1_{i}}=3.680\times 10^3 km/h\\\\speed \quad of\quad moon\quad after\quad collision=v_{1_{f}} = -4.40 \times 10^2 km/h\\\\speed\quad of\quad comet\quad after\quad collision =v_{2_{f}} =5.740 \times 10^3 km/h

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