Respuesta :
Answer : The specific heat of molybdenum metal is, [tex]0.269J/mole^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 molybdenum metal = ?
[tex]c_2[/tex] = specific heat of water = [tex]4.186J/g^oC[/tex]
[tex]m_1[/tex] = mass of molybdenum metal = 237.0 g
[tex]m_2[/tex] = mass of water = 244.0 g
[tex]T_f[/tex] = final temperature of water and metal = [tex]15.30^oC[/tex]
[tex]T_1[/tex] = initial temperature of molybdenum metal = [tex]100.10^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]10.00^oC[/tex]
Now put all the given values in the above formula, we get
[tex]237.0g\times c_1\times (15.30-100.10)^oC=-244.0g\times 4.186J/g^oC\times (15.30-10.00)^oC[/tex]
[tex]c_1=0.269J/g^oC[/tex]
Therefore, the specific heat of molybdenum metal is, [tex]0.269J/mole^oC[/tex]
