Question

Turpentine has a specific heat capacity of 1700j a jar containing 94.8g of turpentine at 15.1 degrees is poured in to a jar containing 152g of turpentine at 27.2 degrees if no thermal energy is lost what is the final temperature of the mixture

Answer 1

Answer: Turpentine has a specific heat capacity of 1700j a jar containing 94.8g of turpentine at 15.1 degrees is poured in to a jar containing 152g of turpentine at 27.2 degrees if no thermal energy is lost what is the final temperature of the mixture

Step-by-step explanation:

Answer 2

howdy!

lets find the answer using the principle of conservation of energy. the equation for this is:

Q1 = Q2

where:

Q1 is the heat gained by the cooler substance (mass m1, specific heat C1, temperature change ΔT1).

Q2 is the heat lost by the warmer substance (mass m2, specific heat C2, temperature change ΔT2).

in this case, we have two portions of turpentine with different temperatures:

m1 = 94.8g, C1 = 1700 J/g°C, T1 = 15.1°C

m2 = 152g, C2 = 1700 J/g°C, T2 = 27.2°C

we need to find the final temperature T.f of the mixture.

first, calculate the heat gained and lost:

Q1 = m1 * C1 * ΔT1

Q2 = m2 * C2 * ΔT2

now, set Q1 = Q2 and solve for ΔT2 (the temperature change of the warmer substance):

ΔT2 = (m1 * C1 * ΔT1) / (m2 * C2)

plug in the values:

ΔT2 = (94.8g * 1700 J/g°C * (T.f - 15.1°C)) / (152g * 1700 J/g°C)

now, solve for T.f:

T.f - 15.1°C = (94.8g * 1700 J/g°C * (T.f - 15.1°C)) / (152g * 1700 J/g°C)

now, isolate T.f:

T.f - 15.1°C = (94.8 / 152) * (T.f - 15.1°C)

now, you can solve for T.f:

(1 / 1) = (94.8 / 152)

T.f - 15.1°C = 0.62368421

T.f = 15.1°C + 0.62368421

T.f ≈ 15.72°C

the final temperature of the mixture is 15.72°C.