Answer:
The final temperature is 71.19 °C
Explanation:
Step 1: Data given
Mass of aluminium = 1.2 kg = 1200 grams
Temperature of aluminium = 20.0 °C
Specific heat of aluminium = 0.900 J/g°C
Mass of water = 1.5 kg = 1500 grams
Temperature of water = 80.0 °C
Specific heat of water = 4.184 J/g°C
Step 2: Calculate the final temperature
heat gained = heat lost
Q(aluminium) = - Q(water)
Q = m*c*ΔT
m(aluminium) * c(aluminium) * ΔT(aluminium) = - m(water) * c(water) * ΔT(water)
⇒ with mass of aluminium = 1200 grams
⇒ with specific heat of aluminium = 0.900 J/g°C
⇒ with ΔT = The change in temperature = T2 - T1 = T2 - 20.0 °C
⇒ with mass of water = 1500 grams
⇒ with specific heat of water = 4.184 J/g°C
⇒ with ΔT = The change in temperature = T2 - T1 = T2 - 80.0°C
1200 * 0.900 * (T2-20.0°C) = -1500 * 4.184 * (T2 - 80.0°C)
1080 * (T2 - 20.0°C) = -6276 * (T2 - 80.0°C)
1080 T2 - 21600 = -6276T2 + 502080
7356T2 = 523680
T2 = 71.19 °C
The final temperature is 71.19 °C
