A 40g sample of polyisoprene, which has a specific heat capacity of f 1.88 J 1.°c, is dropped into an insulated container containing 100g of water at 55C and a constant pressure of 1 atm. The initial temperature of the polyisoprene is 1.5C. 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 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