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When 155 g of water at 20 ᵒC is mixed with 75 g of water at 80 ᵒC, assuming no heat is

lost to the surrounding what is the final temperature of the water. (Specific heat capacity

of water is 4.18 J/°C·g)​

Respuesta :

asuwafohjames

Answer:

39.57 °C

Explanation:

Applying,

Heat gain = Heat lost.

cm'(t₃-t₁) = cm(t₂-t₃)................. Equation 1

Where c = specific heat capacity of water, m = mass of the hotter water, m' = mass of the colder water, t₁ = initial temperature of the colder water, t₂ = initial temperature of the hotter water, t₃ = Final Temperature.

Equation 1 above can futher be simplified to

m'(t₃-t₁) = m(t₂-t₃)................. Equation 2

From the question,

Given: m' = 155 g, m' = 75 g, t₁ = 20 °C, t₂ = 80 °C

Substitute these values into equation 2

155(t₃-20) = 75(80-t₃)

Solve for t₃

155t₃- 3100 = 6000-75t₃

155t₃+75t₃ = 6000+3100

230t₃ = 9100

t₃ = 9100/230

t₃ = 39.57 °C

Lanuel

Assuming no heat is lost to the surrounding, the final temperature of the water is 39.57°C.

Given the following data:

  • Mass of water 1 = 155 g
  • Initial temperature of water 1 = 20°C
  • Mass of water 2 = 75 g
  • Initial temperature of water 2 = 80°C
  • Specific heat capacity of water = 4.18 J/g°C

To determine the final temperature of the water, assuming no heat is lost to the surrounding:

The quantity of heat lost by the water 1 = The quantity of heat gained by the water 2.

[tex]Q_{lost} = Q_{gained}\\\\m_1c\theta = m_2c\theta\\\\m_1\theta = m_2\theta\\\\m_1(T_2 - T_1) = m_2(T_3 - T_2)[/tex]

Substituting the given parameters into the formula, we have;

[tex]155(T_2 -20)=75(80-T_2 )\\\\155T_2 -3100=6000-75T_2\\\\155T_2 +75T_2=6000+3100\\\\230T_2=9100\\\\T_2=\frac{9100}{230}[/tex]

Final temperature = 39.57°C

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