you take 295.5 g of a solid at 30.0 c and let it melt in 425 g of water. the water temperature decreases from 85.1 c to 30.0 c. calculate the heat of fusion of this solid

In this case, since the heat of fusion of a solid substance stands for the energy required to melt it, which is a phase transition from solid to liquid, we can see that the heat lost by water is that gained by the solid, so we can write:

[tex]Q_{solid}=-Q_{w}[/tex]

Thus, by using the water data and its specific heat (4.184), we obtain:

[tex]Q_{solid}=-m_{w}C_{w}(T_f-T_i)\\\\Q_{solid}=-425g*4.184\frac{J}{g\°C}*(30.0-85.1)\°C\\\\ Q_{solid}=97,978.82J=98.0kJ[/tex]

Next, since the heat of fusion of a substance is usually represented in terms of energy per amount of substance, we use the mass of solid to obtain:

[tex]\Delta _{fus}H=\frac{98.0kJ}{295.5g}\\\\ \Delta _{fus}H=0.332\frac{kJ}{g}=332 \frac{J}{g}[/tex]

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Eduard22sly Eduard22sly · Answer 2 · 2021-12-03T12:20:07+01:00

The heat of fusion of the solid is 0.332 KJ/g or 332 J/g

We'll begin by calculating the heat loss by the water. This can be obtained as follow:

Mass of water (M) = 425 g

Initial temperature of water (T₁) = 85.1 °C

Final temperature (T₂) = 30 °C

Change in temperature (ΔT) = T₂ – T₁ = 30 – 85.1 = –55.1 °C

Specific heat capacity of water (C) = 4.184 J/gºC

Heat loss (Q) =?

Q = MCΔT

Q = 425 × 4.184 × –55.1

Q = –97978.82 J

Divide by 1000 to express in KJ

Q = –97978.82 / 1000

Q = –97.98 KJ

Finally, we shall determine the heat fusion of the solid.

Heat loss by water = –97.98 KJ

Heat gained by the solid (Q) = 97.98 KJ

Mass of solid (m) = 295.5 g

Heat of fusion (Hf) =?

Q = m•Hf

97.98 = 295.5 × Hf

Divide both side by 295.5

Hf = 97.98 / 295.5

Hf = 0.332 KJ/g or 332 J/g

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