If the specific heat of ice is 0.5 cal/g degree celsius, how much heat must be released from 5 g of water at 0 degree celsius to change it to ice at -4 degree celsius?
To calculate the amount of heat that must be released, we need to consider the heat required to change the temperature of water from 0°C to 0°C (melting) and then from 0°C to -4°C (cooling).
1. Heat required for melting the ice:
The heat required to change the temperature of 5 g of water from 0°C to 0°C (the melting point) is given by the formula:
Heat = mass * specific heat * change in temperature
Plugging in the values:
Heat = 5 g * 1 cal/g °C * (0°C - 0°C)
Heat = 0 cal
No heat is required to change the temperature of the water from 0°C to 0°C because it's at its melting point.
2. Heat required for cooling the water:
The heat required to change the temperature of 5 g of water from 0°C to -4°C is given by the formula:
Heat = mass * specific heat * change in temperature
Plugging in the values:
Heat = 5 g * 1 cal/g °C * (-4°C - 0°C)
Heat = 5 g * 1 cal/g °C * -4°C
Heat = -20 cal
The negative sign indicates that heat is released during this process.
Therefore, the total heat that must be released from 5 g of water at 0°C to change it to ice at -4°C is 0 cal (for melting) + (-20 cal) (for cooling) = -20 cal.
To calculate the amount of heat required to change the state of water from a liquid to a solid, we need to consider two steps: heating the water to its freezing point and then cooling it to the desired temperature.
Step 1: Heating the water to its freezing point
To heat water from 0°C to its freezing point (0°C), we use the formula:
Q1 = m * c * ΔT
where:
Q1 is the heat transferred,
m is the mass of the water (5 g in this case),
c is the specific heat of water (1 cal/g°C), and
ΔT is the change in temperature (0°C - 0°C = 0°C).
Calculating Q1:
Q1 = (5 g) * (1 cal/g°C) * (0°C - 0°C)
Q1 = 0 cal
Step 2: Cooling the water to -4°C and changing it to ice
To cool the water from 0°C to -4°C, we also use the formula:
Q2 = m * c * ΔT
where:
Q2 is the heat transferred,
m is the mass of the water (5 g in this case),
c is the specific heat of ice (0.5 cal/g°C for this question), and
ΔT is the change in temperature (0°C - (-4°C) = 4°C).
Calculating Q2:
Q2 = (5 g) * (0.5 cal/g°C) * (0°C - (-4°C))
Q2 = (5 g) * (0.5 cal/g°C) * (4°C)
Q2 = 10 cal
To find the total heat required, we add Q1 and Q2:
Total heat required = Q1 + Q2
Total heat required = 0 cal + 10 cal
Total heat required = 10 cal
Thus, 10 calories of heat must be released from 5 grams of water at 0°C to change it to ice at -4°C.