Askiitians Tutor Team
Last Activity: 5 Months ago
To determine the total amount of heat required to convert 10 grams of ice at -5°C to water at 75°C, we need to break down the process into several steps. Each step involves different calculations based on the specific heat capacities and latent heats provided. Let's go through this step-by-step.
Step 1: Heating the Ice from -5°C to 0°C
First, we need to raise the temperature of the ice from -5°C to 0°C. We use the specific heat of ice for this calculation.
- Mass of ice (m): 10 g
- Specific heat of ice (c_ice): 0.5 cal/g°C
- Temperature change (ΔT): 0°C - (-5°C) = 5°C
The heat required (Q1) can be calculated using the formula:
Q1 = m × c_ice × ΔT
Substituting the values:
Q1 = 10 g × 0.5 cal/g°C × 5°C = 25 cal
Step 2: Melting the Ice at 0°C
Next, we need to convert the ice at 0°C to water at 0°C. This involves using the latent heat of fusion.
- Latent heat of fusion (L_f): 80 cal/g
The heat required (Q2) for this phase change is given by:
Q2 = m × L_f
Substituting the values:
Q2 = 10 g × 80 cal/g = 800 cal
Step 3: Heating the Water from 0°C to 75°C
Now, we need to heat the water from 0°C to 75°C. We will use the specific heat of water for this calculation.
- Specific heat of water (c_water): 1 cal/g°C
- Temperature change (ΔT): 75°C - 0°C = 75°C
The heat required (Q3) for this step is calculated as:
Q3 = m × c_water × ΔT
Substituting the values:
Q3 = 10 g × 1 cal/g°C × 75°C = 750 cal
Total Heat Required
Finally, we sum up all the heat calculated in the previous steps to find the total heat required (Q_total):
Q_total = Q1 + Q2 + Q3
Substituting the values:
Q_total = 25 cal + 800 cal + 750 cal = 1575 cal
In summary, the total amount of heat required to convert 10 grams of ice at -5°C to water at 75°C is 1575 calories. This process illustrates the importance of understanding specific heat and latent heat in thermal energy calculations, as each phase change and temperature change requires different amounts of energy.
