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A 38.4−g sample of ethylene glycol, a car radiator coolant, loses 852 J of heat. What was the initial temperature of the ethylene glycol if the final temperature is 32.5°C? (c of ethylene glycol = 2.42 J/g·K)

Respuesta :

aabdulsamir

Answer:

296.33K

Explanation:

Data;

Q = 852J

mass (m) = 38.4g

θ₂ = 32.5°C = 305.5K

θ₁ = ?

c = 2.42 J/g.K

From the equation of heat transfer;

Heat transfer (Q) = MC∇θ

Q = mc(θ₂ - θ₁)

852 = 38.4 * 2.42 * ( 305.5 - θ₁ )

852 = 28389.504 - 92.928θ₁

Collect like-terms

92.928θ₁ = 27537.504

Divide both sides by 92.928 to make θ₁ the subject of formula .

θ₁ = 296.33K

dsdrajlin

A 38.4−g sample of ethylene glycol loses 852 J of heat and its temperature decreases from 41.7 °C to 32.5 °C

A 38.4−g sample of ethylene glycol, a car radiator coolant, loses 852 J of heat., that is, by convention Q = -852 J.

We can calculate the change in the temperature using the following equation.

Q = c × m × ΔT

ΔT = Q / c × m

ΔT = -852 J / (2.42 J/g.°C) × 38.4 g = -9.17 °C

where,

  • Q is the heat lost.
  • c is the specific heat capacity of ethylene glycol (2.42 J/g.K =2.42 J/g.°C).
  • m is the mass of the sample.
  • ΔT is the change in the temperature.

The final temperature is 32.5 °C. Thus, the initial temperature is:

ΔT = T₂ - T₁

T₁ = T₂ - ΔT = 32.5 °C - (-9.17 °C) = 41.7 °C

A 38.4−g sample of ethylene glycol loses 852 J of heat and its temperature decreases from 41.7 °C to 32.5 °C

Learn more about heat here: https://brainly.com/question/16559442

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