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 temperature is 40 deg C. What is the mass of iron?

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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.

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