A 19.0g sample of brass, which has a specific heat capacity of 0.375·J·g-1°C-, is dropped into an insulated container containing 300.0g of water at 20.0°C and a constant pressure of 1atm. The initial temperature of the brass is 81.7°C. Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has the correct number of significant digits.

Answer in degrees C

The equilibrium temperature is 20.35 °C

Explanation:

Step 1: Data given

Mass of brass = 19.0 grams

Specific heat of brass sample = 0.375 J/g°C

Mass of water = 300.0 grams

Initial temperature of water = 20.0°C

Initial temperature of the brass = 81.7 °C

Step 2: Calculate the equilibrium temperature

Heat lost = heat gained

Qbrass = - Qwater

Q = m*c*ΔT

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

⇒ mass of brass sample = 19.0 grams

⇒ c (brass) = 0.375 J/g°C

⇒ ΔT (brass) = The change in temperature = T2 - 81.7 °C

⇒ mass of water = 300.0 grams

⇒ c (water) = 4.184 J/g°C

⇒ ΔT = The change of temperature of water = T2 - 20.0°C

19.0*0.375 * (T2 - 81.7) = - 300.00 * 4.184 * (T2 - 20.0)

7.125 * (T2 - 81.7) = - 1255.2 (T2-20.0)

7.125T2 - 582.1125 = - 1255.2T2 + 25104

1262.325T2 = 25686.1125

T2 = 20.35 °C

The equilibrium temperature is 20.35 °C