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- Homework Statement
- A sphere of 0.047 kg aluminium is placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100 °C. It is then immediately transfered to 0.14 kg copper calorimeter containing 0.25 kg water at 20 °C. The temperature of water rises and attains a steady state at 23 °C. Calculate the specific heat capacity of aluminium.

- Relevant Equations
- Mass of aluminium sphere ##(m_1) = 0.047 kg##

Initial temperature of aluminium sphere##=100 °C##

Final temperature ##= 23 °C##

Change in temperature ##(∆T)=(100 °C-23°C)= 77 °C##

Let specific heat capacity of aluminium be ##s_{Al}##.

The amount of heat lost by the aluminium sphere##= m_1s_{Al}∆T=0.047kg×s_Al ×77°C##

Mass of water ##(m_2) = 0.25 kg##

Mass of calorimeter ##(m_3) = 0.14 kg##

Initial temperature of water and calorimeter##=20 °C##

Final temperature of the mixture ##= 23 °C##

Change in temperature ##(∆T_2) = 23 °C – 20 °C = 3 °C##

Specific heat capacity of water ##(s_w)= 4.18 × 103 J kg–1 K–1##

Specific heat capacity of copper calorimeter ##= 0.386 × 103 J kg–1 K–1##

Heat gained by Calorimeter and Water = Heat Lost by Aluminium sphere

My question :

My understanding is that ##\Delta T = T_f - T_i##

In the case of the calorimeter and water this is consistent with the solution. Yet for the sphere ##\Delta T = T_i - T_f##?

My understanding is that ##\Delta T = T_f - T_i##

In the case of the calorimeter and water this is consistent with the solution. Yet for the sphere ##\Delta T = T_i - T_f##?