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A sample of nickel is heated to [tex]99.8 \, \text{degrees C}[/tex] and placed in a coffee cup calorimeter containing [tex]150.0 \, \text{g}[/tex] of water at [tex]23.5 \, \text{degrees C}[/tex]. After the metal cools, the final temperature of the metal and water mixture is [tex]25.0 \, \text{degrees C}[/tex]. If the specific heat capacity of nickel is [tex]0.444 \, \text{J/(degrees C g)}[/tex], what mass of nickel was originally heated?

Answer :

Answer: The mass of nickel that was originally heated is 28.4 grams.

Explanation:

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]

The equation used to calculate heat released or absorbed follows:

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

[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex] ......(1)

where,

q = heat absorbed or released

[tex]m_1[/tex] = mass of nickel = ?

[tex]m_2[/tex] = mass of water = 150.0 g

[tex]T_{final}[/tex] = final temperature = 25.0°C

[tex]T_1[/tex] = initial temperature of nickel = 99.8°C

[tex]T_2[/tex] = initial temperature of water = 23.5°C

[tex]c_1[/tex] = specific heat of nickel = 0.444 J/g°C

[tex]c_2[/tex] = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

[tex]m_1\times 0.444\times (25.0-99.8)=-[150.0\times 4.186\times (25.0-23.5)][/tex]

[tex]m_1=28.4g[/tex]

Hence, the mass of nickel that was originally heated is 28.4 grams.

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Rewritten by : Jeany

Answer: 28.3 grams

Explanation:

[tex]heat_{absorbed}=heat_{released}[/tex]

As we know that,

[tex]Q=m\times c_1\times \Delta T=m\times c_2\times (T_{final}-T_{initial})[/tex]

[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex] .................(1)

where,

q = heat absorbed or released

[tex]m_1[/tex] = mass of water= 150.0 g

[tex]m_2[/tex] = mass of nickel = ?

[tex]T_{final}[/tex] = final temperature = [tex]25^oC=(273+25)K=298K[/tex]

[tex]T_1[/tex] = temperature of water = [tex]23.5^oC=(273+23.5)K=296.5K[/tex]

[tex]T_2[/tex] = temperature of nickel = [tex]99.8^oC=273+99.8=372.8K[/tex]

[tex]c_1[/tex] = specific heat of water = [tex]4.184J/g^0C[/tex]

[tex]c_2[/tex] = specific heat of nickel= [tex]0.444J/g^0C[/tex]

Now put all the given values in equation (1), we get

[tex]150.0\times 4.184\times (298-296.5)=-[m_2\times 0.444\times (298-372.8)][/tex]

[tex]m_2=28.3g[/tex]

Therefore, the mass of nickel originally heated was 28.3 grams