how much heat must be removed froms 20 g of water at 130 c to make 20 g water at -22 c
To calculate the amount of heat that needs to be removed in this scenario, you can use the formula:
Q = m * C * ΔT
Where:
- Q is the amount of heat transferred (in joules)
- m is the mass of the substance (in grams)
- C is the specific heat capacity of the substance (in J/g°C)
- ΔT is the change in temperature (in °C)
In this case, the substance is water, so its specific heat capacity is 4.184 J/g°C.
Step 1: Calculate the heat required to cool the water from 130°C to 0°C.
Q1 = (20 g) * (4.184 J/g°C) * (0°C - 130°C)
Step 2: Calculate the heat required to change the state of water from 0°C to -22°C.
Q2 = (20 g) * (334 J/g) (since it is the heat required for phase change from liquid to solid)
Step 3: Calculate the heat required to cool the ice from 0°C to -22°C.
Q3 = (20 g) * (2.092 J/g°C) * (-22°C - 0°C)
Q = Q1 + Q2 + Q3
After calculating the values, you can add Q1, Q2, and Q3 together to find the total amount of heat that needs to be removed from the water.
Please note that the negative sign in front of the temperatures indicates that the heat is being removed from the water.