Question

A 4.00 g sample of a metal (specific heat = 0.600 J g-1°C-1 is heated to 75 degrees Celcius and then dropped into 165 g of water in a calorimeter. What is the final temperature of the water if the initial temperature is 28 degrees Celcius? The specific heat capacity of water is 4.184 J/g.°C.

Answer 1

28.16 °C

Step-by-step explanation:

Considering that:-

Heat gain by water = Heat lost by metal

Thus,

`m_(water)* C_(water)* (T_f-T_i)=-m_(metal)* C_(metal)* (T_f-T_i)`

Where, negative sign signifies heat loss

Or,

`m_(water)* C_(water)* (T_f-T_i)=m_(metal)* C_(metal)* (T_i-T_f)`

For water:

Mass = 165 g

Initial temperature = 28 °C

Specific heat of water = 4.184 J/g°C

For metal:

Mass = 4.00 g

Initial temperature = 75 °C

Specific heat of water = 0.600 J/g°C

So,

`165* 4.184* (T_f-28)=4.00* 0.600* (75-T_f)`

`690360\left(T_f-28\right)=2400\left(75-T_f\right)`

`692760T_f=19510080`

`T_f = 28.16\ ^0C`

Hence, the final temperature is 28.16 °C