How many calories must be supplied to 59 g
of ice at −6 C to melt it and raise the temperature of the water to 61 C? The specific heat of ice is 0.49 kcal/kg · K and of water is 1 kcal/kg · K; the heat of fusion of water is 79.7 kcal/kg, and its heat of vaporization is 418 kcal/kg.
Answer in units of kcal
Use two formulas.
Within a phase:
q=mass x specific heat x (Tfinal-Tinitial)
phase change:
mass x heat vap at boiling point
mass x heat fusion at melting point
For example:
at melting point of ice. To change solid ice at zero C to liquid at zero C.
q = 59 g ice x heat fusion - 59 x 80 cal/g = ?
From ice at zero to water 100 C
q = mass x specific heat x(Tfinal-Tinitial) = 59 g x 1 cal/g x (100-0) = ?
Then add all of the q amount to obtain total q the divide by 1000 to convert to Kcal.
To calculate the total number of calories required to melt the ice and raise the temperature of the water, we need to consider two separate processes: melting the ice and raising the temperature of the water.
1. Heat required to melt the ice:
The heat needed to melt 1 gram of ice is called the heat of fusion. In this case, since we have 59 grams of ice, we can use the equation:
Heat required to melt ice = Mass of ice (in grams) × Heat of fusion of water
So, the heat required to melt the ice is:
Heat required to melt ice = 59 g × 79.7 kcal/kg = 4702.3 kcal
2. Heat required to raise the temperature of water:
The heat required to raise the temperature of a substance can be calculated using the specific heat equation:
Heat required = Mass of the substance (in grams) × Specific heat of the substance × Change in temperature
Here, we need to calculate the heat required to raise the temperature of the resulting water from the melting process. The ice is initially at -6°C, and we want to raise the water's temperature to 61°C.
So, the heat required to raise the temperature of the water is:
Heat required to raise temperature of water = Mass of water × Specific heat of water × Change in temperature
We know the mass of water is equal to the mass of ice since the ice melts into water. And the specific heat of water is 1 kcal/kg · K.
Change in temperature = Final temperature - Initial temperature
Change in temperature = 61°C - (-6°C) = 67°C
Heat required to raise temperature of water = 59 g × 1 kcal/kg · K × 67 °C = 3943 kcal
Finally, to find the total number of calories required, we sum the heat required to melt the ice and the heat required to raise the temperature of the water:
Total calories required = Heat required to melt ice + Heat required to raise temperature of water
Total calories required = 4702.3 kcal + 3943 kcal = 8645.3 kcal
Therefore, you must supply 8645.3 kcal of heat to melt the ice and raise the temperature of the resulting water.
To calculate the number of calories required to melt the ice and raise the temperature of the resulting water, we need to consider two steps:
1) Heating the ice from -6 °C to 0 °C (melting point), and
2) Heating the water from 0 °C to 61 °C.
Step 1: Heating the ice to 0 °C
The specific heat of ice is given as 0.49 kcal/kg · K. To find the heat required to heat the ice from -6 °C to 0 °C, we can use the formula:
Q1 = mass × specific heat × change in temperature
Here:
Mass of ice = 59 g = 0.059 kg
Specific heat of ice = 0.49 kcal/kg · K
Change in temperature = 0 °C - (-6 °C) = 6 K
Therefore:
Q1 = 0.059 kg × 0.49 kcal/kg · K × 6 K = 0.17394 kcal
Step 2: Melting the ice and heating the water to 61 °C
The heat of fusion of water is given as 79.7 kcal/kg. This is the heat required to convert ice at 0 °C to water at 0 °C.
Q2 = mass × heat of fusion
Here:
Mass of ice = 59 g = 0.059 kg
Heat of fusion of water = 79.7 kcal/kg
Therefore:
Q2 = 0.059 kg × 79.7 kcal/kg = 4.7143 kcal
Next, we need to heat the resulting water from 0 °C to 61 °C. The specific heat of water is given as 1 kcal/kg · K. To find the heat required to heat the water from 0 °C to 61 °C, we can use the same formula as in Step 1:
Q3 = mass × specific heat × change in temperature
Here:
Mass of water = mass of ice = 0.059 kg
Specific heat of water = 1 kcal/kg · K
Change in temperature = 61 °C - 0 °C = 61 K
Therefore:
Q3 = 0.059 kg × 1 kcal/kg · K × 61 K = 3.599 kcal
Finally, we can calculate the total calories required by adding up the values from each step:
Total Calories = Q1 + Q2 + Q3
Total Calories = 0.17394 kcal + 4.7143 kcal + 3.599 kcal
Total Calories ≈ 8.48724 kcal
Therefore, approximately 8.487 kcal need to be supplied to melt the ice and raise the temperature of the resulting water to 61 °C.