Answer : The initial temperature of water (in
) =
Solution : Given,
Mass of gold = 15.23 g
Mass of water = 28 g
Initial temperature of gold =
Final temperature of gold =
Final temperature of water =
Heat capacity of gold = 
Heat capacity of water = 
The formula used for calorimetry is,
q = m × c × ΔT
where,
q = heat required
m = mass of an element
c = heat capacity
ΔT = change in temperature
In the calorimetry, the energy as heat lost by the system is equal to the gained by the surroundings.
Now the above formula converted and we get

![m_{system}\times c_{system}\times (T_{final}-T_{initial})_{system}= -[m_{surrounding}\times c_{surrounding}\times (T_{final}-T_{initial})_{surrounding}]](https://tex.z-dn.net/?f=m_%7Bsystem%7D%5Ctimes%20c_%7Bsystem%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29_%7Bsystem%7D%3D%20-%5Bm_%7Bsurrounding%7D%5Ctimes%20c_%7Bsurrounding%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29_%7Bsurrounding%7D%5D)
![m_{gold}\times c_{gold}\times (T_{final}-T_{initial})_{gold}= -[m_{water}\times c_{water}\times (T_{final}-T_{initial})_{water}]](https://tex.z-dn.net/?f=m_%7Bgold%7D%5Ctimes%20c_%7Bgold%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29_%7Bgold%7D%3D%20-%5Bm_%7Bwater%7D%5Ctimes%20c_%7Bwater%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29_%7Bwater%7D%5D)
Now put all the given values in this formula, we get
![15.23g\times 12.9J/g^{0}C \times (62^{0}C-53^{0}C )= -[28g\times 4.18J/g^{0}C \times (62^{0}C-T_{\text{ initial of water}})]](https://tex.z-dn.net/?f=15.23g%5Ctimes%2012.9J%2Fg%5E%7B0%7DC%20%5Ctimes%20%2862%5E%7B0%7DC-53%5E%7B0%7DC%20%29%3D%20-%5B28g%5Ctimes%204.18J%2Fg%5E%7B0%7DC%20%5Ctimes%20%2862%5E%7B0%7DC-T_%7B%5Ctext%7B%20initial%20of%20water%7D%7D%29%5D)
By rearranging the terms, we get

Thus the initial temperature of water =