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A piece of aluminum metal at an initial temperature of 97.3°C was placed in a calorimeter containing 49.0g of water at an initial temperature of 22.0°C. The two were allowed to come to thermal equilibrium and the final temperature was 24.1°C. The specific heats of water and aluminum are 4.184 J/g °C and 0.902 J/g °C, respectively. What was the mass of the Al piece that was added?

Respuesta :

Answer : The mass of the Al piece that added was, 6.52 grams.

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 Al = [tex]0.902J/g^oC[/tex]

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

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

[tex]m_2[/tex] = mass of water = 49.0 g

[tex]T_f[/tex] = final temperature of mixture = [tex]24.1^oC[/tex]

[tex]T_1[/tex] = initial temperature of Al = [tex]97.3^oC[/tex]

[tex]T_2[/tex] = initial temperature of water = [tex]22.0^oC[/tex]

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

[tex]m_1\times 0.902J/g^oC\times (24.1-97.3)^oC=-49.0g\times 4.184J/g^oC\times (24.1-22.0)^oC[/tex]

[tex]m_1=6.52g[/tex]

Therefore, the mass of the Al piece that added was, 6.52 grams.

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