The quantity of heat that changes the temperature of a mass of a substance is given by , where is the specific heat capacity of the substance. For example, for , . And for a change of phase, the quantity of heat that changes the phase of a mass is , where is the heat of fusion or heat of vaporization of the substance. For example, for , the heat of fusion is 80 (or 80 ) and the heat of vaporization is 540 (or 540 ).
1.Use these relationships to determine the number of calories to change 1.3 of 0 ice to 0 ice water.
2.Use these relationships to determine the number of calories to change 1.3 of 0 ice water to 1.3 of 100 boiling water.
3.Use these relationships to determine the number of calories to change 1.3 of 100 boiling water to 1.3 of 100 steam.
4. Use these relationships to determine the number of calories to change 1.3 of 0 ice to 1.3 of 100 steam.
To determine the number of calories required to change the state and temperature of the given substances, we need to calculate the heat involved in each step using the provided formulas.
1. Changing 1.3 g of 0°C ice to 0°C ice water:
To calculate the heat required to change the temperature of the ice from 0°C to 0°C water, we use the formula:
Heat = mass * specific heat capacity * change in temperature
Given:
mass = 1.3 g
specific heat capacity (of ice) = ?
change in temperature = 0°C
To find the specific heat capacity of ice, we need to look it up or use a known value. Let's assume the specific heat capacity of ice is 0.5 cal/g°C.
Heat = 1.3 g * 0.5 cal/g°C * 0°C
Heat = 0 calories
Therefore, no calories are required to change 1.3 g of ice at 0°C to ice water at 0°C since there is no change in temperature.
2. Changing 1.3 g of 0°C ice water to 1.3 g of 100°C boiling water:
To calculate the heat required to change the temperature of the ice water from 0°C to 100°C boiling water, we use the formula:
Heat = mass * specific heat capacity * change in temperature
Given:
mass = 1.3 g
specific heat capacity (of water) = ?
change in temperature = 100°C
To find the specific heat capacity of water, we need to look it up or use a known value. The specific heat capacity of water is typically around 1 cal/g°C.
Heat = 1.3 g * 1 cal/g°C * 100°C
Heat = 130 calories
Therefore, it requires 130 calories to change 1.3 g of ice water at 0°C to boiling water at 100°C.
3. Changing 1.3 g of 100°C boiling water to 1.3 g of 100°C steam:
To calculate the heat required to change the state of the boiling water from liquid to steam at the same temperature, we use the formula:
Heat = mass * heat of vaporization
Given:
mass = 1.3 g
heat of vaporization (of water) = 540 cal/g
Heat = 1.3 g * 540 cal/g
Heat = 702 calories
Therefore, it requires 702 calories to change 1.3 g of boiling water at 100°C to steam at 100°C.
4. Changing 1.3 g of 0°C ice to 1.3 g of 100°C steam:
To calculate the total heat required to change the state and temperature of the ice to steam, we add up the heats required for each step.
Heat = Heat to melt ice + Heat to heat ice water to boiling point + Heat to vaporize boiling water
Given:
mass = 1.3 g
specific heat capacity (of ice) = 0.5 cal/g°C
heat of fusion (of ice) = 80 cal/g
specific heat capacity (of water) = 1 cal/g°C
heat of vaporization (of water) = 540 cal/g
Heat = (1.3 g * 0.5 cal/g°C * 100°C) + (1.3 g * 540 cal/g) + (1.3 g * 540 cal/g)
Heat = 65 calories + 702 calories + 702 calories
Heat = 1469 calories
Therefore, it requires 1469 calories to change 1.3 g of ice at 0°C to steam at 100°C.
To determine the number of calories needed for each scenario, we will use the formulas and values provided.
1. Changing 1.3 g of 0°C ice to 0°C ice water:
- The heat required to change the phase from solid ice to liquid water is given by the equation: Q = m × ΔHf
- Here, m represents the mass of the substance and ΔHf represents the heat of fusion of the substance.
- Plugging in the values, Q = 1.3 g × 80 cal/g = 104 calories.
2. Changing 1.3 g of 0°C ice water to 1.3 g of 100°C boiling water:
- The heat required to increase the temperature of a substance can be calculated using the equation: Q = m × c × ΔT
- Here, m represents the mass of the substance, c represents the specific heat capacity of the substance, and ΔT represents the change in temperature.
- Since the ice water is already at 0°C and needs to reach 100°C, ΔT = 100°C - 0°C = 100°C.
- Plugging in the values, Q = 1.3 g × 1 cal/g°C × 100°C = 130 calories.
3. Changing 1.3 g of 100°C boiling water to 1.3 g of 100°C steam:
- Since the boiling water is already at 100°C and needs to change phase to steam, the heat required is the heat of vaporization.
- So, Q = m × ΔHv = 1.3 g × 540 cal/g = 702 calories.
4. Changing 1.3 g of 0°C ice to 1.3 g of 100°C steam:
- First, we need to calculate the heat required to change the temperature of the ice to boiling water.
- Using the heat equation from scenario 2, Q1 = 1.3 g × 1 cal/g°C × (100°C - 0°C) = 130 calories.
- Then, we need to calculate the heat required to change the phase of the boiling water to steam.
- Using the heat equation from scenario 3, Q2 = 1.3 g × 540 cal/g = 702 calories.
- The total heat required is the sum of Q1 and Q2: Q_total = Q1 + Q2 = 130 calories + 702 calories = 832 calories.