Answer:
The equilibrium temperature of the water is 26.7 °C
Explanation:
<u>Step 1:</u> 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
<u>Step 2:</u> 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
