 #1
JC2000
 187
 15
 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 °C23°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##?