A 1.10 kg piece of granite with a specific heat of 0.803 J g-1 °C-1 and a temperature of 87.3 °C is placed into 2.00 L of water at 20.3 °C. When the granite and water come to the same temperature, what will the temperature be?

The final temperature will be 26.7 °C

Explanation:

Step 1: Data given

Mass of the piece granite = 1.10 kg

Specific heat = 0.803 J/g°C

Initial temperature = 87.3 °C

Volume of water = 2.00 L

Initial temperature of water = 20.3°C

Step 2:Calculate the final temperature

Heat lost by granite = heat won by water

Qgranite = - Qwater

Q = m*c*ΔT

m (granite) * c (granite) * ΔT (granite) = - m (water) * c (water) * ΔT (water)

⇒ mass of granite = 1100 grams

⇒ c (granite) = 0.803 J/g°C

⇒ ΔT (granite) = T2 - T1 = T2 - 87.3

⇒ mass of water = 2000 grams

⇒ c (water) = 4.184 J/g°C

⇒ΔT (water) = T2 - 20.3 °C

1100 * 0.803 * (T2 - 87.3) = - 2000 * 4.184 * (T2 - 20.3)

883.3 (T2 - 87.3) = - 8368 (T2 - 20.3)

883.3T2 - 77112.09 = - 8368T2 + 169870.4

883.3T2 + 8368T2 = 169870.4 + 77112.09

9251.3T2 = 246982.49

T2 = 26.7 °C

The final temperature will be 26.7 °C