1.2 kg of aluminum at 20oC is added to 1.5 kg of water at 80oC. After the system reaches thermal equilibrium, what is its final temperature (in oC) ?

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