Question

Need help with chemistry A piece of metal is heated to a temperature of 95.3°C and place in 117 g water which is initially at a temperature of 21.3° C. The final temperature of the metal and water is 32.3° C. The specific heat of the metal is 0.104 cal/g°C. What is the mass of the piece of metal?

Answer 1

We have in this case a transfer of energy from the metal to the water. If we assume that there are no heat losses to the environment, we can say that the heat released by the metal will be the same heat absorbed by the water, it will have the same magnitude but opposite sign. We can represent this with the following equation.

`Q_m=-Q_w`

The subscript 'm' will be for the metal and its properties and the subscript 'w' for water.

Now, the energy of each object will be:

`m_m.Cp_m.\Delta T_m=-m_w.Cp_w.\Delta T_w`

Where,

m is the mass of the object

Cp is the specific heat of the object

dT is the change of temperature

We clear the mass of the metal

`m_m=(-m_w.Cp_w.\Delta T_w)/(Cp_m.\Delta T_m)`

Now, we replace the known data

mw=117g

Cpw=1cal/g°C

dTw=T2-T1=32.3°C-21.3°C=11°C

Cpm=0.104cal/g°C

dTm=T2-T1=32.3°C-95.3°C=-63°C

So, we will have:

Answer: The mass of the piece of metal will be 196g