Question

Question #93a6f

Answer 1

Given
`m_w->"Mass of water"="its volume"xx"density`
`=25cm^3xx1g/(cm^3)=25g`

`t_w->"Initial temperature of water"=23^@C`

`m_(Cd)->"Mass Cadmium"=65.5g`

`t_(Cd)->"Initial temperature of Cd"=100^@C`

`C_w->"Sp.heat capacity of water"=4.184J/(g^@C)`

`C_(Cd)->"Sp.heat capacity of Cd"=0.2311J/(g^@C)`

Let the final temperature of the system be `t^@C`

So by calorimetric principle

`"Heat lost by Cd" = "Heat gained by water"`

`=>m_(Cd)xxC_(Cd)xx(t_(Cd)-t)= m_wxxC_wxx(t-t_w)`

`=>65.5xx0.2311xx(100-t)= 25xx4.184xx(t-23)`

`=>(65.5xx0.2311)/(25xx4.184)xx(100-t)= (t-23)`

`=>0.1447xx(100-t)= (t-23)`

`=>14.47-0.1447t= (t-23)`

`=>1.1447t=37.47`

`=>t=37.47/1.1447~~32.73^@C`