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