Question

I have a few questions in regards to physics, heat, and temperature

1. How much energy must be removed from 5.5 kg of liquid lead at 327oC to produce a block of solid lead at 327C.

2. How many KJ of energy is required to melt exactly 100g of ice initially at -4.00C? Assume no loss of energy to surroundings.

Thank you

Answer 1

Answer on Question #72295 – Physics – Molecular physics – Thermodynamics

Question:

I have a few questions in regards to physics, heat, and temperature

1. How much energy must be removed from $5.5\,\mathrm{kg}$ of liquid lead at 327°C to produce a block of solid lead at 327°C.

2. How many KJ of energy is required to melt exactly 100 g of ice initially at -4.00°C? Assume no loss of energy to surroundings.

Solution:

1. Amount of energy that removes from the system during producing of solid lead from liquid can be calculated from negative value of the enthalpy of fusion:

$$q = -m \Delta H_{fus} = -5.5\,\mathrm{kg} \cdot 2.45 \cdot 10^{4}\,\frac{\mathrm{J}}{\mathrm{kg}} = -134.8\,\mathrm{kJ}$$

2. The full energy of the melting process consists of two energies. There are energy of heating of ice cube from $-4.00\,{}^{\circ}\mathrm{C}$ till the melting point $0\,{}^{\circ}\mathrm{C} - q_1$ and energy of melting of ice cube to liquid water $-q_2$.

$$q = q_1 + q_2$$ $$q_1 = m c \Delta T$$ $$q_2 = m \Delta H_{fus}$$

where $c$ is a heat capacity of water, $\Delta H_{fus}$ is enthalpy of fusion.

$$q = m c \Delta T + m \Delta H_{fus} = m(c \Delta T + \Delta H_{fus})$$ $$q = 100\,\mathrm{g} \cdot \left(4.184\,\frac{\mathrm{J}}{{}{}^{\circ}\mathrm{C} \cdot \mathrm{g}} \cdot (-4.00 - 0)^{{}{}^{\circ}\mathrm{C}} + 334\,\frac{\mathrm{J}}{\mathrm{g}}\right) = 31.7\,\mathrm{kJ}$$

Answer:

during transition of liquid lead to solid block system lost 134.8 kJ; to melt 100 g of ice the system need 31.7 kJ.

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