Question

A piece of gold with a mass of 15.23g and an initial temp of 53 Celsius was dropped into a calorimeter containing 28g of water, the final temp of the metal and water in the calorimeter was 62 what is the initial temp

Answer 1

Answer : The initial temperature of water (in $^{0}celcius$) = $77.108^{0}C$  

Solution : Given,

Mass of gold = 15.23 g

Mass of water = 28 g

Initial temperature of gold = $53^{0}C$  

Final temperature of gold = $62^{0}C$

Final temperature of water = $62^{0}C$

Heat capacity of gold = $12.9J/g^{0}C$

Heat capacity of water = $4.18J/g^{0}C$

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

$q_{system}= - q_{surrounding}$

$m_{system}\times c_{system}\times (T_{final}-T_{initial})_{system}= -[m_{surrounding}\times c_{surrounding}\times (T_{final}-T_{initial})_{surrounding}]$

$m_{gold}\times c_{gold}\times (T_{final}-T_{initial})_{gold}= -[m_{water}\times c_{water}\times (T_{final}-T_{initial})_{water}]$

Now put all the given values in this formula, we get

$15.23g\times 12.9J/g^{0}C \times (62^{0}C-53^{0}C )= -[28g\times 4.18J/g^{0}C \times (62^{0}C-T_{\text{ initial of water}})]$

By rearranging the terms, we get

$T_{\text{ initial water}=77.108^{0}C$

Thus the initial temperature of water = $77.108^{0}C$