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

Answer:

28.16 °C

Explanation:

Considering that:-

Heat gain by water = Heat lost by metal

Thus,  

$m_{water}\times C_{water}\times (T_f-T_i)=-m_{metal}\times C_{metal}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,  

$m_{water}\times C_{water}\times (T_f-T_i)=m_{metal}\times C_{metal}\times (T_i-T_f)$

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,  

$165\times 4.184\times (T_f-28)=4.00\times 0.600\times (75-T_f)$

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

$692760T_f=19510080$

$T_f = 28.16\ ^0C$

Hence, the final temperature is 28.16 °C