How long would it take a 1.00 x 10^3 W heating element to melt 2.00kg of ice at -20 deg C, assuming all energy is absorbed by the ice?
10^3 * time in seconds t = Joules in
so
10^3 t = 2 kg*2000 J/(kg deg)*(20 deg) + 2 kg (334*10^3 J/kg)
10^3 t = 80* 10^3 J + 668 *10^3 J
t = 748 s or about 12.5 minutes
To find out how long it would take a 1.00 x 10^3 W heating element to melt 2.00 kg of ice at -20°C, we need to calculate the heat required to melt the ice, and then divide it by the power of the heating element.
Here are the step-by-step calculations:
Step 1: Calculate the heat required to raise the temperature of the ice from -20°C to 0°C.
- The specific heat capacity of ice is 2.09 J/g°C.
- The mass of the ice is 2.00 kg.
Therefore, the heat required is:
Q1 = mass x specific heat capacity x change in temperature
= 2.00 kg x 2.09 J/g°C x (0 - (-20)°C)
= 2.00 kg x 2.09 J/g°C x 20°C
= 83.6 x 10^3 J
Step 2: Calculate the heat required to melt the ice.
- The heat of fusion of ice is 334 J/g.
Therefore, the heat required is:
Q2 = mass x heat of fusion
= 2.00 kg x 334 J/g
= 668 x 10^3 J
Step 3: Calculate the total heat required.
- Add the heat required to raise the temperature (Q1) and the heat required to melt the ice (Q2).
Therefore, the total heat required is:
Q_total = Q1 + Q2
= 83.6 x 10^3 J + 668 x 10^3 J
= 751.6 x 10^3 J
Step 4: Calculate the time required.
- Divide the total heat required (Q_total) by the power of the heating element.
Therefore, the time required is:
t = Q_total / power
= (751.6 x 10^3 J) / (1.00 x 10^3 W)
= 751.6 seconds
Converting the time to minutes:
1 minute = 60 seconds
Therefore, the total time required is:
t = 751.6 seconds / 60
= 12.53 minutes
So, it would take approximately 12.53 minutes for the 1.00 x 10^3 W heating element to melt 2.00 kg of ice at -20°C, assuming all energy is absorbed by the ice.
To determine how long it would take a heating element to melt a certain amount of ice, we need to consider the energy required to melt the ice.
First, we calculate the energy required to heat the ice from -20°C to 0°C using the specific heat capacity of ice. The specific heat capacity of ice is 2.09 J/g°C.
Energy required to heat the ice = (mass of ice) x (specific heat capacity of ice) x (change in temperature)
= 2.00 kg x 2.09 J/g°C x (0°C - (-20°C))
= 2.00 kg x 2.09 J/g°C x 20°C
= 83.6 kJ
Next, we calculate the energy required to melt the ice. The specific heat of fusion for ice is 333.55 J/g.
Energy required to melt the ice = (mass of ice) x (specific heat of fusion)
= 2.00 kg x 333.55 J/g
= 667.10 kJ
Now, we add the energy required for heating and melting the ice:
Total energy required = Energy required to heat the ice + Energy required to melt the ice
= 83.6 kJ + 667.10 kJ
= 750.7 kJ
Given that the heating element has a power of 1.00 x 10^3 W (watts), we can calculate the time required to provide this amount of energy using the formula:
Time = Total energy required / Power
Time = 750.7 kJ / 1.00 x 10^3 W
= 750.7 s
So, it would take the heating element approximately 750.7 seconds to melt 2.00 kg of ice at -20°C, assuming all the energy is absorbed by the ice.