Question

A) Suppose 0.395 kg of water initially at 41.5°C is poured into a 0.300 kg glass beaker having a temperature of 25.0°C. A 0.500 kg block of aluminum at 37.0°C is placed in the water, and the system insulated. Calculate the final equilibrium temperature of the system. °C

B) A 21.5 kg gold bar at 27.0°C is placed in a large, insulated 0.800 kg glass container at 15.0°C with 2.00 kg of water at 25.0°C. Calculate the final equilibrium temperature. °C

Answer 1

ONE QUESTION AT A TIME PLEASE

A) Using energy transfer equation

Q_water + Q_aluminium + Q_{glass beaker} = 0

C_water = 4190 J/kg ^oC , C_aluminium = 900 J/kg ^oC , C_{glass beaker} = 837 J/kg ^oC

m_waterC_water ( T - T_water) = 0.395*4190 ( T - 41.5) = 1655.5 (T - 41.5) --- (1)

m_AluminiumC_aluminium( T - T_aluminium) = 450( T - 37) ------------(2)

m_{glass beaker}C_{glass beaker(} T - T_{glass beaker}) = 251.1(T - 25) ------------- (3)

Now put this in main equilibrium equation, we get

2356.15T = 91630.75

T = 38.89^oC