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ladessa [460]
3 years ago
9

Para que sirve la caida libre

Physics
1 answer:
masya89 [10]3 years ago
4 0
Eh? i don’t understand huhu
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If you have 240.0 mL of water at 25.00 °C and add 100.0 mL of water at 95.00 °C, what is the final temperature of the mixture? U
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<u>Answer:</u> The final temperature of the mixture is 45.6°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

  • <u>When volume is 240.0 mL</u>

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{240.0mL}\\\\\text{Mass of water}=(1g/mL\times 240.0mL)=240g

  • <u>When volume is 100.0 mL</u>

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{100.0mL}\\\\\text{Mass of water}=(1g/mL\times 100.0mL)=100g

When two water solutions at different temperature are mixed, the amount of heat released by water present at higher temperature will be equal to the amount of heat absorbed by water present at lower temperature..

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of water solution 1 = 240 g

m_2 = mass of water solution 2 = 100 g

T_{final} = final temperature = ?°C

T_1 = initial temperature of water solution 1 = 25°C

T_2 = initial temperature of water solution 2 = 95°C

c = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

240\times 4.186\times (T_{final}-25)=-[100\times 4.186\times (T_{final}-95)]\\\\T_{final}=45.6^oC

Hence, the final temperature of the mixture is 45.6°C

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3 years ago
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