Answer:
The equilibrium temperature of the water is 46.85 °C
Explanation:
Step 1: Data given
Mass of sample of polyisoprene = 40.0 grams
Specific heat capacity of polyisoprene = 1.88 J/g°C
Mass of water = 100 grams
Temperature of water = 55.0 °C
Pressure = 1 atm
The initial temperature of the polyisoprene is 1.5 °C
Step 2: Calculate the equilibrium temperature of the water
Heat lost = Heat gained
Qlost = - Qgained
Qwater = -Qpolyisoprene
Q = m*c*ΔT
m(water)*c(water)*ΔT(water) = -m(polyisoprene) * c(polyisoprene) * ΔT(polyisoprene)
⇒with m(water) = the mass of water = 100 grams
⇒with c(water) = the specific heat of water = 4.184 J/g°C
⇒with ΔT = the change of temeprature = T2 - T1 = T2 - 55.0°C
⇒with m(polyisoprene) = the mass of polyisoprene = 40.0 grams
⇒with c(polyisoprene) = the specific heat of polyisoprene = 1.88 J/g°C
⇒with ΔT(polyisoprene) =the change of temperature = T2 -T1 = T2 - 1.5 °C
100 * 4.184 * (T2 - 55.0) = -40.0 * 1.88 * (T2 - 1.5)
418.4(T2 - 55.0) = -75.2(T2 - 1.5)
418.4 T2 - 23012 = -75.2T2+ 112.8
493.6 T2 = 23124.8
T2 = 46.85 °C
The equilibrium temperature of the water is 46.85 °C
