Respuesta :

Explanation:

In the first experiment there is 11 g of water    ( 48 - 37 g = 11g)

  specific heat of water is  4.186 J / (g C)

    for 11 g and 400 J :

            400 J / (11 g C)  = 4.186  J / (g C)    shows C = 8.69 degrees C

So the final temp would be   60 + 8.69 C = 68.69° C

in the second experiment , there is   72.5 - 40.5 g of the solid = 32 g

 and it takes  12 736 J to melt this 32 g solid object

   the heat of fusion is then   12 736 J / 32 g = 398 J/g

   ( Note that the melting point temp is not needed )

semsee45

Answer:

63) Final temperature of water = 68.7 °C

64) Heat of fusion of the solid =  398 J/g

Explanation:

Practice 63

To determine the final temperature of the water, we can use the formula for heat transfer:

[tex]Q = cm\rm\Delta T[/tex]

where:

  • Q is the heat absorbed (400 J in this case).
  • c is the specific heat capacity of the substance (water ≈ 4.186 J/g°C)
  • m is the mass of the substance (water in the calorimeter).
  • ΔT is the change in temperature.

The mass of water (m) can be calculated as the difference in mass between the calorimeter + water and the mass of the calorimeter:

[tex]m = \text{(Mass of calorimeter + water)} - \text{Mass of calorimeter}[/tex]

[tex]m = \rm 48.0\;g-37.0\;g[/tex]

[tex]m = \rm 11.0\;g[/tex]

Substitute the values into the heat transfer formula and solve for the change in temperature (ΔT):

[tex]\rm 400\;J = 4.186\;J/g^{\circ}C \cdot 11.0\;g \cdot \Delta T[/tex]

[tex]\rm \Delta T=\dfrac{400\;J}{4.186\;J/g^{\circ}C \cdot 11.0\;g}[/tex]

[tex]\rm \Delta T=8.6869652087...^{\circ}C[/tex]

Now, to find the final temperature, simply add the change in temperature (ΔT) to the initial temperature:

[tex]\rm Final\;temperature=Initial\;temperature+ \Delta T[/tex]

[tex]\rm Final\;temperature=60.0\;^{\circ}C+8.6869652087...\;^{\circ}C[/tex]

[tex]\rm Final\;temperature=68.6869652087...\;^{\circ}C[/tex]

[tex]\rm Final\;temperature=68.7\;^{\circ}C\;(3\;s.f.)[/tex]

Therefore, the final temperature of the water would be approximately 68.7 °C (rounded to three significant figures).

[tex]\hrulefill[/tex]

Practice 64

The heat of fusion is determined by dividing the energy absorbed or released during a phase transition by the mass of the substance undergoing either melting or freezing.

To calculate the heat of fusion of the solid, we can use the heat of fusion formula:

[tex]\Delta H_f = \dfrac{q}{m}[/tex]

where:

  • [tex]\Delta H_f \;\text{is the heat of fusion.}[/tex]
  • [tex]\text{$q$ is the heat absorbed by the solid during melting.}[/tex]
  • [tex]\text{$m$ is the mass of the substance melted.}[/tex]

(Note that the melting point of the solid does not directly affect the calculation of the heat of fusion when using this formula).

The mass of the solid (m) can be calculated as the difference in mass between the calorimeter + solid and the mass of the calorimeter:

[tex]m = \text{(Mass of calorimeter + solid)} - \text{Mass of calorimeter}[/tex]

[tex]m = \rm 72.5\;g-40.5\;g[/tex]

[tex]m = \rm 32.0\;g[/tex]

Now, substitute the given values into the heat of fusion formula:

[tex]\Delta H_f = \rm \dfrac{12736\;J}{32.0\;g}[/tex]

[tex]\Delta H_f =\rm 398\;J/g[/tex]

Therefore, the heat of fusion of the solid is 398 J/g.

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