Answer : The correct option is, (a) Gold
Solution :
\(Q_{absorbed}=Q_{released}\)
As we know that,
\(Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})\)
\(m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]\) .................(1)
where,
\(m_1\) = mass of metal = 10 g
\(m_2\) = mass of added water = 30 g
\(T_{final}\) = final temperature = ?
\(T_1\) = temperature of metal = \(65^oC\)
\(T_2\) = temperature of added water = \(258^oC\)
\(c_2\) = specific heat of water = \(4.184J/g^oC\)
\(c_1\) = specific heat of metal
Now we have to calculate the final temperature for all the given metals.
(a) For gold : given,
\(c_1\) = specific heat of metal = \(0.129J/g^oC\)
\(10g\times 0.129J/g^oC\times (T_{final}-65^oC)=-[30g\times 4.184J/g^oC\times (T_{final}-25^oC)]\)
\(T_{final}=25.4^oC\)
Change in temperature of water = \((25.4-25)^oC=0.4^oC\)
(b) For silver : given,
\(c_1\) = specific heat of metal = \(0.24J/g^oC\)
\(10g\times 0.24J/g^oC\times (T_{final}-65^oC)=-[30g\times 4.184J/g^oC\times (T_{final}-25^oC)]\)
\(T_{final}=25.7^oC\)
Change in temperature of water = \((25.7-25)^oC=0.7^oC\)
(c) For copper : given,
\(c_1\) = specific heat of metal = \(0.385J/g^oC\)
\(10g\times 0.385J/g^oC\times (T_{final}-65^oC)=-[30g\times 4.184J/g^oC\times (T_{final}-25^oC)]\)
\(T_{final}=26.19^oC\)
Change in temperature of water = \((26.19-25)^oC=1.19^oC\)
(d) For aluminium : given,
\(c_1\) = specific heat of metal = \(0.902J/g^oC\)
\(10g\times 0.902J/g^oC\times (T_{final}-65^oC)=-[30g\times 4.184J/g^oC\times (T_{final}-25^oC)]\)
\(T_{final}=27.6^oC\)
Change in temperature of water = \((27.6-25)^oC=2.6^oC\)
Therefore, the least temperature rise for water result in, Gold.
