Question

QUICK!!!

The specific heat of mercury is 0.138 J/g Celsius. If 452 g of mercury at 85.0 Celsius are placed in 145 g of water at 23.0 Celsius, what will be the final temperature for both the mercury and the water?

Answer 1

Answer: Thus the final temperature for both the mercury and the water is
`28.8^0C`

Step-by-step explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

`heat_(released)=heat_(absorbed)`

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

`-[m_1* c_1* (T_(final)-T_1)]=[m_2* c_2* (T_(final)-T_2)]`

Q = heat absorbed or released

`m_1` = mass of mercury= 452 g

`m_2`= mass of water = 145 g

`T_(final)` = final temperature = ?

`T_1` = temperature of mercury =
`85.0^0C`

`T_2` = temperature of water =
`23.0^0C`

`c_1` = specific heat of mercury =
`0.138J/g^0C`

`c_2` = specific heat of water=
`4.184J/g^0C`

Now put all the given values in equation (1), we get

`-[452* 0.138* (T_f-85.0)^0C]=145* 4.184* (T_f-23.0)^0C`

`T_f=28.8^0C`

Thus the final temperature for both the mercury and the water is
`28.8^0C`