Question

A 230g lead ball at a temperature of 81.3⁰C is placed in a light calorimeter containing 158g of water at 22.3⁰C Find the equilibrium temperature of the system

Answer 1

The equilibrium temperature of the system, given that lead at 81.3 °C is placed in a calorimeter containing water at 22.3 °C, is 24.8 °C

How to calculate the equilibrium temperature of the sysytem?

The equilibrium temperature of the system can calculated by applying the conservation of thermal energy. This is shown below:

• Mass of lead (M) = 230 g
• Temperature of lead (T) = 81.3 °C
• Specific heat capacity of lead (C) = 0.129 J/g°C
• Mass of water (Mᵥᵥ) = 158 g
• Temperature of water (Tᵥᵥ) = 22.3 °C
• Specific heat capacity of water (Cᵥᵥ) = 4.184 J/g°C
• Equilibrium temperature of system (Tₑ) =?

Heat released by lead = Head gained by water

`MC(T\ -\ T_e) = M_wC_w(T_e\ -\ T_w)\\\\230\ *\ 0.129\ *\ (81.3\ -\ T_e) = 158\ *\ 4.184\ *\ (T_e\ -\ 22.3)\\\\2412.171 -\ 29.67T_e = 661.072T_e\ -\ 14741.9056`

`2412.171\ +\ 14741.9056 = 661.072T_e\ +\ 29.67T_e\\\\17154.0766 = 690.742T_e\\\\T_e = (17154.0766)/(690.742) \\\\T_e = 24.8\ \textdegree C`