Question

A calorimeter is used to measure the specific heat of an object. The water bath has an initial temperature of 23.2 ° C. The object with a temperature of 67.8 ° C is placed in the Beaker.After thermal equilibrium is established, the temperature of the water bath is 25.6 ° CWhat is the specific heat of the object (Lead)? We give ; Specific heat of water ce = 4185J.Kg and Density = 1000kg / m

Answer 1

Consider that the mass of water bath is 0.700 kg and the mass of object is 0.200 kg.

The heat required to change the temperature of water is,

`Q_w=m_wC_w(T-T_w)`

The heat required to change the temprature of object is,

`Q_o=m_oC_o(T_o-T)`

Assume that no heat is lost. Therefore,

`Q_w=Q_o`
`m_wC_w(T-T_w)=m_oC_o(T_o-T)`
`C_o=\frac{m_{w_{}}C_w(T-T_w)}{m_o(T_o-T)}`

Substitute the known values,

`\begin{gathered} C_o=\frac{(0.700\text{ kg)(4185 J/K}.\text{g)(25.6}^(\circ)\text{C-23.2}^(\circ)\text{C)}}{(0.200\text{ kg)(67.8}^(\circ)\text{C-}25.6^(\circ)C)} \\ =\frac{(0.700\text{ kg)(4185 J/K.g)(2.4}^(\circ)\text{C)}}{(0.200\text{ kg)(42.2}^(\circ)\text{C})}(\frac{1\text{ kg}}{1000\text{ g}}) \\ \approx0.833\text{ J/kg.K} \end{gathered}`

Therefore, the specific heat of lead is 0.833 J/kg.K