contestada


The heat of a reaction may be found with the equation q=mcΔT.

A car engine is in danger of being damaged from wear and stress if not properly cooled by the radiator fluid. The 2.51 kg sample of radiator fluid is heated from 23.8°C to 205°C by this process. The specific heat capacity of radiator fluid, containing a 50/50 mixture of ethylene glycol and water, is 3.41 J/(g*°C) Assuming there is no heat lost to the surroundings, calculate the heat released by the engine.

A) -1961 kJ
B) -1551 kJ
C) 1751 kJ
D) 2041 kJ

Respuesta :

znk

Answer:

B) -1551 kJ  

Explanation:

There are two heat flows in this question.

Heat released by engine + heat absorbed by water = 0

                   q₁                    +                      q₂                = 0

                   q₁                    +                   mCΔT            = 0

Data:

  m = 2.51 kg

  C = 3.41 J°C⁻¹g⁻¹

T_i =   23.8 °C

T_f =205     °C  

Calculations:

(a) ΔT

ΔT = 205 °C - 23.8 °C = 181.2 °C

(b) q₂

q₂ = mCΔT = 2510 g × 3.41 J·°C⁻¹g⁻¹ × 181.2 °C = 1.551× 10⁶ J = 1551 kJ

(c) q₁

q₁ + 1551 kJ = 0

q₁ = -1551 kJ

Otras preguntas