Question

An unknown material has a normal melting/freezing point of -27.8 °C, and the liquid phase has a specific heat capacity of 157 J/(kg C°). One-tenth of a kilogram of the solid at -27.8 °C is put into a 0.169-kg aluminum calorimeter cup that contains 0.101 kg of glycerin. The temperature of the cup and the glycerin is initially 26.6 °C. All the unknown material melts, and the final temperature at equilibrium is 18.3 °C. The calorimeter neither loses energy to nor gains energy from the external environment. What is the latent heat of fusion of the unknown material?

Answer 1

Answer on Question #76832, Physics / Molecular Physics | Thermodynamics

An unknown material has a normal melting/freezing point of $-27.8{}^{\circ}\mathrm{C}$ , and the liquid phase has a specific heat capacity of $157\mathrm{J} / (\mathrm{kg}\mathrm{C}^{*})$ . One-tenth of a kilogram of the solid at $-27.8{}^{\circ}\mathrm{C}$ is put into a 0.169-kg aluminum calorimeter cup that contains $0.101\mathrm{kg}$ of glycerin. The temperature of the cup and the glycerin is initially $26.6{}^{\circ}\mathrm{C}$ . All the unknown material melts, and the final temperature at equilibrium is $18.3{}^{\circ}\mathrm{C}$ . The calorimeter neither loses energy to nor gains energy from the external environment. What is the latent heat of fusion of the unknown material?

Solution

When an unknown material is put into aluminum calorimeter cup that contains glycerin a heat exchange takes place: an unknown material absorbs heat that aluminum and glycerin give off until the system comes to the state of heat equilibrium at temperature $18.3{}^{\circ}\mathrm{C}$ . The value of heat absorbed is equal to the value of heat given off.

$$Q _ {\text {a b s o r b e d}} = Q _ {\text {g i v e n o f f}}$$

1. $Q_{\text{absorbed}} = Q_1 + Q_2$

Where

a) $Q_{1}$ – the heat of phase change of an unknown material:

$$Q _ {1} = m \cdot L _ {f},$$

m- mass of an unknown material, $m = 0.1\mathrm{kg}$

$L_{f} =$ latent heat of fusion.

$$Q _ {1} = 0. 1 \cdot L _ {f}$$

b) $Q_{2}$ – heat that is required to raise the temperature of $0.1 \, \text{kg}$ of unknown material from

-27.8°C to 18.3°C.

$$Q _ {2} = c m \left(T _ {2} - T _ {1}\right)$$

c- specific heat capacity of an unknown material, $c = 157 \, \text{J/(kg C}^*)$ ,

m-mass of an unknown material, $m = 0.1$ kg

$$T _ {2} = 1 8. 3 {}^ {\circ} \mathrm {C}$$ $$T _ {1} = - 2 7. 8 {}^ {\circ} \mathrm {C}$$ $$Q _ {2} = 1 5 7 \cdot 0. 1 (1 8. 3 - (- 2 7. 8)) = 7 2 3. 7 7 (\mathrm {J}).$$

Then $Q_{\text{absorbed}} = 0.1 \cdot L_f + 723.77$ (J)

2. $Q_{\text{given off}} = Q_{\text{Al}} + Q_{\text{gly}}$

Where

a) $Q_{Al}$ – the heat that $0.169 \, \text{kg}$ of aluminum calorimeter cup gives off when the temperature decreases from $26.6{}^{\circ} \text{C}$ to $18.3{}^{\circ} \text{C}$ .

$$Q _ {A l} = c m \left(T _ {2} - T _ {1}\right)$$

c- specific heat capacity of aluminum, c =900 J/(kg C*),

m- mass of aluminum calorimeter cup, $m = 0.169\mathrm{kg}$

$$T _ {2} = 1 8. 3 {}^ {\circ} \mathrm {C}$$ $$T _ {1} = 2 6. 6 {}^ {\circ} \mathrm {C}$$ $$Q _ {A l} = 9 0 0 \cdot 0. 1 6 9 \cdot (1 8. 3 - 2 6. 6) = - 1 2 6 2. 4 3 (\mathrm {J}).$$

b) $Q_{\mathrm{gly}}$ – the heat that $0.101 \, \mathrm{kg}$ of glycerin gives off when the temperature decreases from $26.6{}^{\circ} \mathrm{C}$ to $18.3{}^{\circ} \mathrm{C}$ .

$$Q _ {g l y} = c m \left(T _ {2} - T _ {1}\right)$$

c- specific heat capacity of glycerin, c =2410 J/(kg C*),

m- mass of glycerin, $m = 0.101\mathrm{kg}$

$$T _ {2} = 1 8. 3 {}^ {\circ} \mathrm {C}$$ $$T _ {1} = 2 6. 6 {}^ {\circ} \mathrm {C}$$ $$Q _ {A l} = 2 4 1 0 \cdot 0. 1 0 1 \cdot (1 8. 3 - 2 6. 6) = - 2 0 2 0. 3 (\mathrm {J}).$$

Then $Q_{\text{given off}} = -1262.43 - 2020.3 = -3282.73$ (J).

Minus before value of $Q_{\text{given off}}$ shows that the system (aluminum and glycerin) looses heat.

Find $L_{f}$ from the equation:

$$\left| Q _ {\text {a b s o r b e d}} \right| = \left| Q _ {\text {g i v e n o f f}} \right|$$ $$0. 1 \cdot L _ {f} + 7 2 3. 7 7 = 3 2 8 2. 7 3,$$ $$L _ {f} = 2 5 5 8 9. 6 \mathrm {J / k g} = 2 5. 6 \mathrm {k J / k g}.$$

Answer: $25.6\mathrm{kJ / kg}$

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