How much heat is necessary to change 30 g of ice at -20°C into water at 20°C?
c(ice)= 2 060 J/kg•degr
c(water)= 4180 J/kg•degr
λ = 335000 J/kg
Q=Q1+Q2+Q3
Q1=m•c(ice) •ΔT= 0.03•2060 • 20 =1236 J.
Q2=λm=335000•0.03=10050 J
Q3 = m•c(water) •ΔT=0.03•4180•20=2508 J
Q=1236+10050+2508 =13794 J
To determine the amount of heat required to change a given substance from one state to another, we need to understand the concept of heat transfer and use the specific heat capacity formula.
First, we need to break down the process into two steps:
1. Heating the ice from -20°C to 0°C (phase change from solid to liquid).
2. Heating the water from 0°C to 20°C (temperature change within the liquid state).
Step 1: Heating the ice to 0°C
To calculate the heat required to raise the temperature of the ice, we use the formula:
Q = m * c * ΔT
Where:
Q = heat transferred (in Joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature
The specific heat capacity of ice is approximately 2.09 J/g°C. Therefore, to raise the temperature of the ice from -20°C to 0°C:
Q1 = 30 g * 2.09 J/g°C * (0°C - (-20°C))
Q1 = 30 g * 2.09 J/g°C * 20°C
Q1 = 1254 J
Step 2: Heating the water to 20°C
To calculate the heat required to heat the water from 0°C to 20°C, we again use the formula:
Q = m * c * ΔT
The specific heat capacity of water is approximately 4.18 J/g°C. Therefore, to raise the temperature of the water from 0°C to 20°C:
Q2 = 30 g * 4.18 J/g°C * (20°C - 0°C)
Q2 = 30 g * 4.18 J/g°C * 20°C
Q2 = 2510 J
Now, to find the total heat required for the whole process, we sum up Q1 and Q2:
Total heat = Q1 + Q2
Total heat = 1254 J + 2510 J
Total heat = 3764 J
Therefore, approximately 3764 Joules of heat is required to change 30 g of ice at -20°C into water at 20°C.
To determine the amount of heat necessary to change ice at -20°C into water at 20°C, we need to consider three steps:
1. Heating the ice from -20°C to 0°C to melt it
2. Adding heat to the water at 0°C to bring it to 100°C
3. Heating the water from 100°C to 20°C
Each step requires a specific amount of heat.
Step 1: Heat required to melt the ice
The heat required to melt the ice can be calculated using the formula:
Q = m * L_f
where:
Q = heat energy (in Joules)
m = mass of the ice (in grams)
L_f = latent heat of fusion (in Joules/gram)
The latent heat of fusion for ice is approximately 334 J/g.
Q1 = 30 g * 334 J/g = 10,020 J
Step 2: Heat required to go from 0°C to 100°C
The heat required to raise the temperature of water is calculated using the formula:
Q = m * c * ΔT
where:
Q = heat energy (in Joules)
m = mass of the water (in grams)
c = specific heat capacity of water (approximately 4.18 J/g°C)
ΔT = change in temperature (in °C)
In this case, the change in temperature is from 0°C to 100°C.
Q2 = 30 g * 4.18 J/g°C * (100°C - 0°C) = 12,540 J
Step 3: Heat required to go from 100°C to 20°C
Using the same formula as in step 2, we calculate the heat required for this step.
Q3 = 30 g * 4.18 J/g°C * (20°C - 100°C) = -18,864 J (negative sign indicates heat is released)
Total heat required:
To find the total heat required to change the ice at -20°C into water at 20°C, we sum up the three steps:
Total heat = Q1 + Q2 + Q3
Total heat = 10,020 J + 12,540 J - 18,864 J
Total heat = 3,696 J
Therefore, approximately 3,696 Joules of heat is needed to change 30 grams of ice at -20°C into water at 20°C.