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

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

Step-by-step 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`

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_(absorbed)=heat_(released)`

As we know that,

`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)]` .................(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* c* (75-34)=[115.4* 4.184* (34-22)]`

`c=0.58J/g^0C`

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