Question

A certain substance has a specific heat of 0.355 J/g °C for the solid, a specific heat of 0.552 J/g °C for the liquid, and a melting point of 843 °C. A 14.3 gram sample of the substance required 4.35 kJ of energy to change its temperature from 825 °C to 853 °C. (a) What was the heat of fusion for the substance in cal/g.


(b) If energy was being added to the substance at a rate of 15 J/s, how many minutes would it take for 6.25 g of the substance to melt?

Answer 1

Answer on Question #83457, Chemistry / General Chemistry

A certain substance has a specific heat of 0.355 J/g °C for the solid, a specific heat of 0.552 J/g °C for the liquid, and a melting point of 843 °C. A 14.3 gram sample of the substance required 4.35 kJ of energy to change its temperature from 825 °C to 853 °C. (a) What was the heat of fusion for the substance in cal/g.

(b) If energy was being added to the substance at a rate of 15 J/s, how many minutes would it take for 6.25 g of the substance to melt?

Solution

a) $Q_{\text{total}} = c_{\text{s}} m(T_{\text{mp}} - T_{1}) + \lambda m + c_{\text{liq}} m(T_{2} - T_{\text{mp}})$

$$\lambda = \frac{Q_{\text{tot}} - c_{\text{s}} m(T_{\text{mp}} - T_{1}) - c_{\text{liq}} m(T_{2} - T_{\text{mp}})}{m}$$ $$\lambda = \frac{4350 - 0.355 \times 14.3 \times (843 - 825) - 0.552 \times 14.3 \times (853 - 843)}{14.3} = 292.3 \, (\mathrm{J/g}) = 69.8 \, (\mathrm{cal/g})$$

b) $Q = \lambda m = 292.3 \times 6.25 = 1827 \, (\mathrm{J})$

$$t = \frac{Q}{V} = \frac{1827}{15} = 121.8 \, (\mathrm{s}) \cong 2 \, \text{min} \, 2 \, \mathrm{s}$$

Answer

The heat of fusion for the substance is 69.8 (cal/g).

It would take 2 min 2 s to melt 6.25 g of the substance.

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