The heat that must be absorbed by ice of mass 500 g at – 10°C to take it to water at 20°C is (Specific heat of Ice is 2.2 kJ/kg K, Specific heat of water is 4.2 kJ/kg K and Latent heat of fusion of ice is 300 kJ/kg)
203 kJ
Heat Absorption Calculation Explained
To determine the total heat absorbed by the ice, we need to consider three distinct stages of heat absorption as the substance transitions from ice at –10°C to water at 20°C.
Here are the given values:
- Mass of ice (\(m\)) = 500 g = 0.5 kg (since 1 kg = 1000 g)
- Initial temperature of ice = –10°C
- Final temperature of water = 20°C
- Specific heat of ice (\(c_{ice}\)) = 2.2 kJ/kg K
- Specific heat of water (\(c_{water}\)) = 4.2 kJ/kg K
- Latent heat of fusion of ice (\(L_f\)) = 300 kJ/kg
Let's break down the process into stages:
Stage 1: Heat Absorbed by Ice from –10°C to 0°C
In this stage, the temperature of the ice increases without changing its phase. The heat absorbed (\(Q_1\)) can be calculated using the formula:
\(Q = m \times c \times \Delta T\)
- Change in temperature (\(\Delta T_{ice}\)) = Final temperature – Initial temperature = 0°C – (–10°C) = 10°C.
- Since 1°C change is equal to 1 K change, \(\Delta T_{ice}\) = 10 K.
So, \(Q_1 = 0.5 \text{ kg} \times 2.2 \text{ kJ/kg K} \times 10 \text{ K}\)
\(Q_1 = 11 \text{ kJ}\)
Stage 2: Heat Absorbed to Melt Ice at 0°C into Water at 0°C
This stage involves a phase change from ice to water at a constant temperature (0°C). The heat absorbed (\(Q_2\)) is known as the latent heat of fusion and is calculated using the formula:
\(Q = m \times L_f\)
So, \(Q_2 = 0.5 \text{ kg} \times 300 \text{ kJ/kg}\)
\(Q_2 = 150 \text{ kJ}\)
Stage 3: Heat Absorbed by Water from 0°C to 20°C
In this final stage, the temperature of the water increases. The heat absorbed (\(Q_3\)) is calculated using the specific heat of water:
\(Q = m \times c \times \Delta T\)
- Change in temperature (\(\Delta T_{water}\)) = Final temperature – Initial temperature = 20°C – 0°C = 20°C.
- \(\Delta T_{water}\) = 20 K.
So, \(Q_3 = 0.5 \text{ kg} \times 4.2 \text{ kJ/kg K} \times 20 \text{ K}\)
\(Q_3 = 42 \text{ kJ}\)
Total Heat Absorbed Calculation
The total heat absorbed (\(Q_{total}\)) is the sum of the heat absorbed in all three stages:
\(Q_{total} = Q_1 + Q_2 + Q_3\)
\(Q_{total} = 11 \text{ kJ} + 150 \text{ kJ} + 42 \text{ kJ}\)
\(Q_{total} = 203 \text{ kJ}\)
Therefore, the total heat that must be absorbed by the ice is 203 kJ.
Here's a summary table of the calculations:
| Stage | Process | Formula | Calculation | Heat Absorbed (kJ) |
|---|---|---|---|---|
| 1 | Ice heating from –10°C to 0°C | \(Q_1 = m \times c_{ice} \times \Delta T_{ice}\) | \(0.5 \text{ kg} \times 2.2 \text{ kJ/kg K} \times 10 \text{ K}\) | 11 |
| 2 | Ice melting at 0°C into water at 0°C | \(Q_2 = m \times L_f\) | \(0.5 \text{ kg} \times 300 \text{ kJ/kg}\) | 150 |
| 3 | Water heating from 0°C to 20°C | \(Q_3 = m \times c_{water} \times \Delta T_{water}\) | \(0.5 \text{ kg} \times 4.2 \text{ kJ/kg K} \times 20 \text{ K}\) | 42 |
| Total Heat Absorbed | 203 | |||
Important Questions from The Perfect Gas
The quantity of heat required to raise the temperature of unit mass of a material by one degree centigrade is called
The amount of heat required for converting one kilogram of a solid completely into liquid is called:
2 kg of substance receives 500 kJ and undergoes a temperature change from 100°C to 200°C. The average specific heat of substance during the process will be
The general law for the expansion or compression of gases is:
Amount of energy required to raise the temperature of a substance of 1 kg mass by 1°C is called
