Understanding the concepts of specific heat and latent heat is crucial for analyzing how materials respond to heat. Specific heat refers to the amount of heat required to change the temperature of a substance, while latent heat pertains to the heat absorbed or released during a phase change without a temperature change. These two concepts are often combined in problems involving heating processes.
The phases of matter—solid, liquid, and gas—are determined by temperature. For example, water (H2O) exists as ice (solid) at low temperatures, as liquid water at moderate temperatures, and as steam (gas) at high temperatures. When heat is added to a substance, it can either increase the temperature or cause a phase change, but not both simultaneously. This principle is illustrated in a temperature versus heat graph.
When heating ice, the temperature rises until it reaches the melting point (0°C or 273 K). At this point, the temperature remains constant while the ice transforms into water. The heat absorbed during this phase change is calculated using the formula:
\( q = mL_f \)
where \( L_f \) is the latent heat of fusion. Conversely, when heating water to its boiling point (100°C or 373 K), the temperature remains constant while the water changes to steam. The heat absorbed during this phase change is given by:
\( q = mL_v \)
where \( L_v \) is the latent heat of vaporization.
To solve problems involving these concepts, it is helpful to visualize the process using a temperature versus heat graph. For example, to calculate the heat required to raise the temperature of 0.4 kg of water from 80°C to 100°C, the equation used is:
\( q = mc\Delta T \)
where \( c \) is the specific heat capacity of water (approximately 4186 J/kg·°C) and \( \Delta T \) is the change in temperature (20°C in this case). Plugging in the values gives:
\( q = 0.4 \, \text{kg} \times 4186 \, \text{J/kg·°C} \times 20 \, \text{°C} = 33,500 \, \text{J} \)
For boiling half of the water at 100°C, the heat required is calculated using the latent heat of vaporization:
\( q = mL_v \)
For 0.2 kg of water, using \( L_v \approx 2.256 \times 10^6 \, \text{J/kg} \), the calculation yields:
\( q = 0.2 \, \text{kg} \times 2.256 \times 10^6 \, \text{J/kg} = 451,200 \, \text{J} \)
Finally, to completely boil all the water starting from 80°C, the total heat required combines both the heating to 100°C and the boiling process:
\( q_{\text{total}} = mc\Delta T + mL_v \)
Substituting the values gives:
\( q_{\text{total}} = 0.4 \, \text{kg} \times 4186 \, \text{J/kg·°C} \times 20 \, \text{°C} + 0.4 \, \text{kg} \times 2.256 \times 10^6 \, \text{J/kg} = 936,000 \, \text{J} \)
By mastering these calculations and understanding the underlying principles, students can effectively tackle a variety of problems involving heat transfer and phase changes.
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