Question

A 25.0g sample of brass, which has a specific heat capacity of 0.375·J·g−1°C−1, is dropped into an insulated container containing 250.0g of water at 25.0°C and a constant pressure of 1atm. The initial temperature of the brass is 96.7°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 3 significant digits.

Answer 1

The equilibrium temperature of water is 25.6 °C

Step-by-step explanation:

Step 1: Data given

Mass of the sample of brass = 25.0 grams

The specific heat capacity = 0.375 J/g°C

Mass of water = 250.0 grams

Temperature of water = 25.0 °C

The initial temperature of the brass is 96.7°C

Step 2: Calculate the equilibrium temperature

Heat lost = heat gained

Q(sample) = -Q(water)

Q = m*C* ΔT

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

⇒m(sample) = the mass of the sample of brass = 25.0 grams

⇒with c(sample) =The specific heat capacity = 0.375 J/g°C

⇒with ΔT = the change of temperature = T2 - T1 =T2 - 96.7 °C

⇒with m(water) = the mass of the water = 250.0 grams

⇒with c(water) = the specific heat capacity = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = T2 - 25.0°C

25.0 * 0.375 * (T2 - 96.7) = - 250.0 * 4.184 J/g°C * (T2 - 25.0°C)

9.375T2 - 906.56 = -1046T2 + 26150

9.375T2 + 1046T2 = 26150 + 906.56

1055.375T2 = 27056.26

T2 = 25.6 °C

The equilibrium temperature of water is 25.6 °C