A sphere of aluminium of0.047 kg placed for sufficient time in avessel containing boiling water, so that thesphere is at 100 °C. It is then immediatelytransfered to 0.14 kg copper calorimetercontaining 0.25 kg of water at 20 ° C. Thetemperature of water rises and attains asteady state at 23 °C. Calculate the specificheat capacity of aluminium

To solve this problem, we will use the principle of conservation of energy, which states that the heat lost by the hot object (the aluminium sphere) is equal to the heat gained by the cold objects (the copper calorimeter and the water).

First, let's calculate the heat gained by the water and the calorimeter:

The heat gained by the water (Q_water) can be calculated using the formula Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. For water, the specific heat capacity is approximately 4.18 J/g°C.

Q_water = m_water * c_water * ΔT_water Q_water = 0.25 kg * 4.18 kJ/kg°C * (23°C - 20°C) Q_water = 3.14 kJ

The heat gained by the calorimeter (Q_calorimeter) can also be calculated using the formula Q = mcΔT. For copper, the specific heat capacity is approximately 0.39 J/g°C.

Q_calorimeter = m_calorimeter * c_calorimeter * ΔT_calorimeter Q_calorimeter = 0.14 kg * 0.39 kJ/kg°C * (23°C - 20°C) Q_calorimeter = 0.17 kJ

The total heat gained by the water and the calorimeter is Q_total = Q_water + Q_calorimeter = 3.14 kJ + 0.17 kJ = 3.31 kJ.

Next, let's calculate the specific heat capacity of aluminium:

The heat lost by the aluminium sphere (Q_aluminium) is equal to the total heat gained by the water and the calorimeter. We can rearrange the formula Q = mcΔT to solve for the specific heat capacity c:

c_aluminium = Q_aluminium / (m_aluminium * ΔT_aluminium) c_aluminium = 3.31 kJ / (0.047 kg * (100°C - 23°C)) c_aluminium = 0.91 kJ/kg°C

Therefore, the specific heat capacity of aluminium is approximately 0.91 kJ/kg°C.