A48.0g sample of quartz, which has a specific heat capacity of 0.730·j·g−1°c−1, is dropped into an insulated container containing 300.0g of water at 25.0°c and a constant pressure of 1atm. the initial temperature of the quartz is 88.6°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.

The equilibrium temperature of the water is 26.7 °C

Explanation:

Step 1: Data given

Mass of the sample quartz = 48.0 grams

Specific heat capacity of the sample = 0.730 J/g°C

Initial temperature of the sample = 88.6°C

Mass of the water = 300.0 grams

Initial temperature = 25.0°C

Specific heat capacity of water = 4.184 J/g°C

Step 2: Calculate final temperature

Qlost = -Qgained

Qquartz = - Qwater

Q =m*c*ΔT

Q = m(quartz)*c(quartz)*ΔT(quartz) = -m(water) * c(water) * ΔT(water)

⇒ mass of the quartz = 48.0 grams

⇒ c(quartz) = the specific heat capacity of quartz = 0.730 J/g°C

⇒ ΔT(quartz) = The change of temperature of the sample = T2 -88.6 °C

⇒ mass of water = 300.0 grams

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

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

48.0 * 0.730 * (T2-88.6) -300.0 * 4.184 *(T2 - 25.0)

35.04(T2-88.6) = -1255.2 (T2-25)

35.04T2 -3104.544 = -1255.2T2 + 31380

1290.24T2 = 34484.544

T2 = 26.7 °C

The equilibrium temperature of the water is 26.7 °C