A piece of iron at 215 deg C and a 20 kg piece of copper at 140 deg C are placed in 25 kg of water at 10 deg C. At thermal equilibrium, the...

When the warmer objects (pieces of iron and copper) are placed in contact with the colder object (water), there will be a transfer of heat so that the iron and copper will cool off and the water will heat up. 

According to the law of the conservation of energy,

`Q_(hot) + Q_(cold) = 0` .

Here,

`Q_(hot)` is the heat leaving the warmer objects (it will have a negative value), and

`Q_(cold)` is the heat acquired by the...

When the warmer objects (pieces of iron and copper) are placed in contact with the colder object (water), there will be a transfer of heat so that the iron and copper will cool off and the water will heat up. 

According to the law of the conservation of energy,

`Q_(hot) + Q_(cold) = 0` .

Here,

`Q_(hot)` is the heat leaving the warmer objects (it will have a negative value), and

`Q_(cold)` is the heat acquired by the colder object as the result.

In this case,

`Q_(hot) = c_im_i(T_e - T_(ii)) + c_cm_c(T_e-T_(ic))`

and `Q_(cold) = c_wm_w(T_e-T_(iw))`

Here, c's denote the specific heat of iron, copper and water, m's denote the masses and `T_i`

 - initial temperature of the objects.

 `T_e`

is the equilibrium temperature.

Plugging in the values for given quantities (masses are in kilograms), and the table values for specific heat (in J/(kg*C)), we get

`448*m_i*(40-215) + 387*20*(40-140) + 4186*25*(40-10) = 0`

From here,

`-78400*m_i - 774000+3139500 = 0 `

Solving for the mass of iron results in

`m_i = 30.17` kg 

The mass of iron is 30.17 kilograms.