How many minutes would be required for a 310.0 W immersion heater to heat 202.0 g of water from 22.0°C to 95.6°C?
Joules of energy required
... 202 * (95.6 - 22.0) * 4.18
divide by heater wattage
... 1 watt = 1 Joule per second
divide by 60 ... sec to min
Well, let me put on my clown hat and calculate this for you! But don't worry, it won't be as dry as a desert.
First, we need to find the amount of heat required to heat the water. We can use the specific heat capacity equation, which is Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
m = 202.0 g
ΔT = (95.6°C - 22.0°C) = 73.6°C
Now, the specific heat capacity of water is approximately 4.18 J/g°C (don't worry, I won't give you a chemistry lecture!).
So, Q = (202.0 g) x (4.18 J/g°C) x (73.6°C)
Now, let's calculate the time it takes to supply this amount of heat using the power equation: P = W/t, where P is power, W is work, and t is time.
Given:
P = 310.0 W (power)
Now, we know that W = Q (the work is equal to the amount of heat transferred).
So, W = (202.0 g) x (4.18 J/g°C) x (73.6°C)
And, P = W/t
Rearranging the equation to find t, we get t = W/P.
Plug in the values, and you should get your answer in minutes. Just a tiny bit longer than the time it takes for a joke to sink in!
To calculate the time required to heat the water, we can use the formula:
Q = m * c * ΔT
Where:
Q = heat energy required (in Joules)
m = mass of water (in grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (final temperature - initial temperature)
First, let's calculate the change in temperature:
ΔT = 95.6°C - 22.0°C
ΔT = 73.6°C
Next, we'll calculate the heat energy required:
Q = 202.0 g * 4.18 J/g°C * 73.6°C
Q = 60717.92 Joules
Since the power of the immersion heater is given in watts, we need to convert the heat energy to watts:
1 Joule = 1 watt * 1 second
60717.92 J = 60717.92 watt-seconds
Since power is equal to energy divided by time:
Power = Energy / Time
We can rearrange the equation to solve for time:
Time = Energy / Power
Time = 60717.92 watt-seconds / 310.0 W
Time = 196.15 seconds
However, the question asks for the time in minutes, so let's convert seconds to minutes:
Time = 196.15 seconds * (1 minute / 60 seconds)
Time = 3.27 minutes
Therefore, it would take approximately 3.27 minutes to heat 202.0 g of water from 22.0°C to 95.6°C using a 310.0 W immersion heater.
To determine the number of minutes required for the immersion heater to heat the water, we'll need to use the formula:
Q = mcΔT
Where:
Q = heat energy in Joules
m = mass of water in grams
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature
First, let's calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 95.6°C - 22.0°C
ΔT = 73.6°C
Next, let's convert the mass of water from grams to kilograms:
mass = 202.0 g ÷ 1000 = 0.202 kg
Now, we can calculate the heat energy:
Q = mcΔT
Q = 0.202 kg × 4.18 J/g°C × 73.6°C
Note: We convert grams to kilograms to match the units of specific heat capacity.
Q ≈ 61.60 kJ (kilojoules)
Finally, we can calculate the time required using the power formula:
Power (P) = Energy (Q) / Time (t)
The power of the immersion heater is given as 310.0 W. We'll convert the heat energy from kilojoules to joules:
Q = 61.60 kJ × 1000 = 61,600 J
Now we can solve for time:
P = Q / t
310.0 W = 61,600 J / t
To find t, rearrange the equation:
t = Q / P
t = 61,600 J / 310.0 W
t ≈ 198.71 seconds
To convert seconds to minutes, divide by 60:
t ≈ 198.71 s / 60 ≈ 3.31 minutes
Therefore, it would take approximately 3.31 minutes for the 310.0 W immersion heater to heat the 202.0 g of water from 22.0°C to 95.6°C.