Answer : The correct option is, (a) Gold
Solution :
![Q_{absorbed}=Q_{released}](https://tex.z-dn.net/?f=Q_%7Babsorbed%7D%3DQ_%7Breleased%7D)
As we know that,
![Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})](https://tex.z-dn.net/?f=Q%3Dm%5Ctimes%20c%5Ctimes%20%5CDelta%20T%3Dm%5Ctimes%20c%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29)
.................(1)
where,
= mass of metal = 10 g
= mass of added water = 30 g
= final temperature = ?
= temperature of metal = ![65^oC](https://tex.z-dn.net/?f=65%5EoC)
= temperature of added water = ![258^oC](https://tex.z-dn.net/?f=258%5EoC)
= specific heat of water = ![4.184J/g^oC](https://tex.z-dn.net/?f=4.184J%2Fg%5EoC)
= specific heat of metal
Now we have to calculate the final temperature for all the given metals.
(a) For gold : given,
= specific heat of metal = ![0.129J/g^oC](https://tex.z-dn.net/?f=0.129J%2Fg%5EoC)
![10g\times 0.129J/g^oC\times (T_{final}-65^oC)=-[30g\times 4.184J/g^oC\times (T_{final}-25^oC)]](https://tex.z-dn.net/?f=10g%5Ctimes%200.129J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-65%5EoC%29%3D-%5B30g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-25%5EoC%29%5D)
![T_{final}=25.4^oC](https://tex.z-dn.net/?f=T_%7Bfinal%7D%3D25.4%5EoC)
Change in temperature of water = ![(25.4-25)^oC=0.4^oC](https://tex.z-dn.net/?f=%2825.4-25%29%5EoC%3D0.4%5EoC)
(b) For silver : given,
= specific heat of metal = ![0.24J/g^oC](https://tex.z-dn.net/?f=0.24J%2Fg%5EoC)
![10g\times 0.24J/g^oC\times (T_{final}-65^oC)=-[30g\times 4.184J/g^oC\times (T_{final}-25^oC)]](https://tex.z-dn.net/?f=10g%5Ctimes%200.24J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-65%5EoC%29%3D-%5B30g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-25%5EoC%29%5D)
![T_{final}=25.7^oC](https://tex.z-dn.net/?f=T_%7Bfinal%7D%3D25.7%5EoC)
Change in temperature of water = ![(25.7-25)^oC=0.7^oC](https://tex.z-dn.net/?f=%2825.7-25%29%5EoC%3D0.7%5EoC)
(c) For copper : given,
= specific heat of metal = ![0.385J/g^oC](https://tex.z-dn.net/?f=0.385J%2Fg%5EoC)
![10g\times 0.385J/g^oC\times (T_{final}-65^oC)=-[30g\times 4.184J/g^oC\times (T_{final}-25^oC)]](https://tex.z-dn.net/?f=10g%5Ctimes%200.385J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-65%5EoC%29%3D-%5B30g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-25%5EoC%29%5D)
![T_{final}=26.19^oC](https://tex.z-dn.net/?f=T_%7Bfinal%7D%3D26.19%5EoC)
Change in temperature of water = ![(26.19-25)^oC=1.19^oC](https://tex.z-dn.net/?f=%2826.19-25%29%5EoC%3D1.19%5EoC)
(d) For aluminium : given,
= specific heat of metal = ![0.902J/g^oC](https://tex.z-dn.net/?f=0.902J%2Fg%5EoC)
![10g\times 0.902J/g^oC\times (T_{final}-65^oC)=-[30g\times 4.184J/g^oC\times (T_{final}-25^oC)]](https://tex.z-dn.net/?f=10g%5Ctimes%200.902J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-65%5EoC%29%3D-%5B30g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-25%5EoC%29%5D)
![T_{final}=27.6^oC](https://tex.z-dn.net/?f=T_%7Bfinal%7D%3D27.6%5EoC)
Change in temperature of water = ![(27.6-25)^oC=2.6^oC](https://tex.z-dn.net/?f=%2827.6-25%29%5EoC%3D2.6%5EoC)
Therefore, the least temperature rise for water result in, Gold.