Respuesta :
Answer : The initial temperature of water (in [tex]^{0}celcius[/tex]) = [tex]77.108^{0}C[/tex]
Solution : Given,
Mass of gold = 15.23 g
Mass of water = 28 g
Initial temperature of gold = [tex]53^{0}C[/tex]
Final temperature of gold = [tex]62^{0}C[/tex]
Final temperature of water = [tex]62^{0}C[/tex]
Heat capacity of gold = [tex]12.9J/g^{0}C[/tex]
Heat capacity of water = [tex]4.18J/g^{0}C[/tex]
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
[tex]q_{system}= - q_{surrounding}[/tex]
[tex]m_{system}\times c_{system}\times (T_{final}-T_{initial})_{system}= -[m_{surrounding}\times c_{surrounding}\times (T_{final}-T_{initial})_{surrounding}][/tex]
[tex]m_{gold}\times c_{gold}\times (T_{final}-T_{initial})_{gold}= -[m_{water}\times c_{water}\times (T_{final}-T_{initial})_{water}][/tex]
Now put all the given values in this formula, we get
[tex]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}})][/tex]
By rearranging the terms, we get
[tex]T_{\text{ initial water}=77.108^{0}C[/tex]
Thus the initial temperature of water = [tex]77.108^{0}C[/tex]
