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A 89.1 g piece of copper (specific heat 0.38 J/g･°C) is heated and then placed into 400.0 g of water initially at 20.7°C. The water increases in temperature to 22.2°C. What is the initial temperature of the copper? (The specific heat of water is 4.18 J/g･°C).

Respuesta :

Answer:

The initial temperature of copper is 96.3 °C

Explanation:

Specific heat capacity is the energy needed to raise the temperature of one gram of material by one degree celsius. The energy absorbed or released by a material during temperature change can be calculated by the below formula:

[tex]Q=mc(T_{2}-T_{1})[/tex] , where:

Q = energy (J)

m = mass (g)

c = specific heat capacity (J/g·°C)

[tex]T_{1}[/tex] = initial temperature (°C)

[tex]T_{2}[/tex] = final temperature (°C)

In an ideal situation, it can be assumed that all the energy lost by the piece of copper is gained by the water, resulting in its temperature rise and that there is no change of mass/state for either material. Thus, the equation can be written as below:

[tex]+Q_{c}=-Q_{w}[/tex]

[tex]+m_{c}c_{c}(T_{c2}-T_{c1})=-m_{w}c_{w}(T_{w2}-T_{w1})[/tex]

It can also be assumed that the final temperature of both the copper and water are the same. Thus substituting below values in above equation will give:

[tex]m_{w}[/tex] = 400 g

[tex]c_{w}[/tex] = 4.18 J/g·°C

[tex]T_{w1}[/tex] = 20.7 °C

[tex]T_{w2}[/tex] = 22.2 °C

[tex]m_{c}[/tex] = 89.1 g

[tex]c_{c}[/tex] = 0.38 J/g·°C

[tex]T_{c1}[/tex] = ? °C

[tex]T_{c2}[/tex] = 22.2 °C

[tex]+89.1*0.38*(22.2-T_{c1})=-400*4.18(22.2-20.7)[/tex]

Solving for [tex]T_{c1}[/tex] gives:

[tex]T_{c1}[/tex] = 96.3 °C

adioabiola

Here, we are required to determine the initial temperature of the copper.

The initial temperature of the copper is;

T(c1) = 96.27 °C

The energy required to raise the temperature of one gram of a material by one degree celsius is termed the Specific Heat Capacity of that material.

Mathematically, we have;

Q=mc{T(2) -T(1)}

Q=mc{T(2) -T(1)} where:

  • Q=mc{T(2) -T(1)} where:Q = energy (J)

  • Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)

  • Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)

  • Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)

  • Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)

Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:

Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:+Q(c) =−Q(w)

Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:+Q(c) =−Q(w)

m(c) × c(c) × (T2(c) - T1(c)) = -{m(w) × c(w) × (T2(w) - T1(w))}

However, the final temperature of the copper piece and water can be assumed to be equal

i.e. T2(c) = T1(w)

  • m(w) = 400g

  • c(w) = 4.18 J/g·°C

  • T (w1) = 20.7 °C

  • T(w2) = 22.2 °C

  • m(c) = 89.1 g

  • c(c) = 0.38 J/g·°C

  • T(c1) = ? °C

  • T(c2) = 22.2 °C

+89.1*0.38*(22.2-T(c1)) = -400*4.18(22.2-20.7)

(22.2−T(c1) ) = −2508/33.858

(22.2−T(c1) ) = -74.07

T(c1) = 74.07 + 22.2

T(c1) = 96.27 °C.

Therefore, the initial temperature of the copper is; T(c1) = 96.27 °C

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