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)* c_(system)* (T_(final)-T_(initial))_(system)= -[m_(surrounding)* c_(surrounding)* (T_(final)-T_(initial))_(surrounding)]`

`m_(gold)* c_(gold)* (T_(final)-T_(initial))_(gold)= -[m_(water)* c_(water)* (T_(final)-T_(initial))_(water)]`

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

`15.23g* 12.9J/g^(0)C * (62^(0)C-53^(0)C )= -[28g* 4.18J/g^(0)C * (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`