Question icon
Grade 12th passThermal Physics

the amount of heat required to convert 10 gm of ic at -5 c water at 75 c latent heat fusion of ice =80 cal/gm latent heated of vapouration of water=540 cal/gm specific heat of water =1 cal/gm/c and specific heat of ice =0.5 cal/gm/c

Profile image of siva
9 Years agoGrade 12th pass
Answers icon

1 Answer

Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer1 Year 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.