Answer:
The equilibrium temperature is 73.34 °C
Explanation:
Step 1: Data given
Mass of aluminium = 100.0 grams
Initial temperature of aluminium = 23.0 °C
Specific heat of aluminium = 0.900 J/g°C
Mass of water = 50.0 grams
Initial temperature of water = 95.0 °C
Specific heat of water = 4.184 J/g°C
Step 2: Calculate equilibrium temperature
Heat gained = heat lost
Q = m*c*ΔT
Qaluminium = - Qwater
m(aluminium) * c(aluminium) * ΔT(aluminium) = -m(water) * c(water) * ΔT(water)
⇒mass aluminium = 100.0 grams
⇒ c(aluminium) = specific heat aluminium = 0.900 J/g°C
⇒ΔT = Change in temperature of aluminium = T2 - T1 = T2 - 23.0 °
⇒mass water = 50.0 grams
⇒ cwater) = specific heat ater = 4.184 J/g°C
⇒ΔT = Change in temperature of water = T2 - T1 = T2 - 95.0 °
100.0g * 0.900 J/g°C * (T2- 23.0°C) = - 50.0g * 4.184J/g°C *(T2- 95.0°C)
90*(T2-23.0°C) = -209.2(T2- 95.0°C)
90T2 -2070 = -209.2T2 + 19874
299.2T2 = 21944
T2 = 73.34 °C
The equilibrium temperature is 73.34 °C
