Question

A 245.7g sample of metal at 75.0℃ was placed in 115.4g of water at 22.0℃. The final temperature of the water and metal was 34.0℃. If no heat was lost to the surroundings, what is the specific heat of the metal? (specific heat of water = 4.184 J/g℃)

Answer 1

Answer:

The answer to your question is:  Cp = 0.6373 J/g°C

Explanation:

Sample

mass = 245.7 g

T1 = 75°C

T2 = 34°

Cp = ?

Water

Mass = 115.4 g

T1 = 22°C

T2 = 34°

Cp = 4.184 J/g°C

Heat of metal = mCp(T2 - T1) = 245.7Cp (34 - 75) = -9090.9Cp

Heat of water =                       = 115.4(4.184)(34 - 22) = -5794

                       Heat of metal = heat of water

                       -9090.9Cp = -5794

                       Cp = -5794 / -9090.9

                       Cp = 0.6373 J/g°C   specific heat of metal

Answer 2

Answer: The specific heat of the metal is $0.58J/g^0C$

Explanation:

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

$heat_{absorbed}=heat_{released}$

As we know that,  

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

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$         .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 245.7 g

$m_2$ = mass of water = 115.4 g

$T_{final}$ = final temperature = $34^0C$

$T_1$ = temperature of metal = $75^oC$

$T_2$ = temperature of water = $22^oC$

$c_1$ = specific heat of metal = ?

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

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

$-245.7\times c\times (75-34)=[115.4\times 4.184\times (34-22)]$

$c=0.58J/g^0C$

Therefore, the specific heat of the metal is $0.58J/g^0C$