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 $28.8^0C$

Explanation:

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

$heat_{released}=heat_{absorbed}$

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$-[m_1\times c_1\times (T_{final}-T_1)]=[m_2\times c_2\times (T_{final}-T_2)]$

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

$-[452\times 0.138\times (T_f-85.0)^0C]=145\times 4.184\times (T_f-23.0)^0C$

$T_f=28.8^0C$

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