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

Respuesta :

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.

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