A1.10 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

the correct answer is:

a. formation of precipitation