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A 100.0g sample of lead is heated to 100.0 oC (Celsius) and is placed in a coffee cup calorimeter containing 150. g of water at 25.0 oC. After the metal cools, the final temperature of the metal and the water is 26.4 oC. Calculate the specific heat capacity of lead from these experimental data, assuming that no heat escapes to the surroundings or is transferred to the calorimeter. Specific heat of water = 4.184 J/g oC atomic mass Pb = 207.2 g/mol Calculate your answer in J/g oC with 3 significant figures, but enter the answer without units.

Respuesta :

Answer : The specific heat capacity of lead is, [tex]0.119J/g^oC[/tex]

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

[tex]q_1=-q_2[/tex]

[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]

where,

[tex]c_1[/tex] = specific heat of lead = ?

[tex]c_2[/tex] = specific heat of water = [tex]4.184J/g^oC[/tex]

[tex]m_1[/tex] = mass of lead = 100.0 g

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

[tex]T_f[/tex] = final temperature = [tex]26.4^oC[/tex]

[tex]T_1[/tex] = initial temperature of lead = [tex]100.0^oC[/tex]

[tex]T_2[/tex] = initial temperature of water = [tex]25^oC[/tex]

Now put all the given values in the above formula, we get

[tex]100.0g\times c_1\times (26.4-100.0)^oC=-150.0g\times 4.184J/g^oC\times (26.4-25)^oC[/tex]

[tex]c_1=0.119J/g^oC[/tex]

Therefore, the specific heat capacity of lead is, [tex]0.119J/g^oC[/tex]

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