My apology. I posted the previous question in correctly. Here is the question again:
To bring 100cc of water from 20C to 75C, in 1min 25 sec, with 650Watt of energy input, what is the temperature required?
To solve this problem, we can use the formula for calculating the amount of heat energy required to raise the temperature of a substance:
Q = mcΔT
Where:
Q = heat energy (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
In this case, we're trying to find the temperature, so let's rearrange the formula:
ΔT = Q / (mc)
Now, we need to determine the values of m and c. The specific heat capacity of water is approximately 4.18 J/g°C, and we know that we want to raise the temperature of 100cc (which is equivalent to 100g, assuming water's density is 1g/mL) of water from an initial temperature of 20°C to the desired temperature.
ΔT = Q / (100g * 4.18 J/g°C)
Now, let's calculate the amount of heat energy using the power equation:
P = ΔQ / Δt
Where:
P = power input (in watts)
ΔQ = change in heat energy (in joules)
Δt = change in time (in seconds)
We are given that the power input is 650 watts and the time for heating is 1 minute 25 seconds, which can be converted to 85 seconds.
650W = ΔQ / 85s
Rearranging the equation:
ΔQ = 650W * 85s
Now, let's substitute the value of ΔQ into our earlier equation:
ΔT = (650W * 85s) / (100g * 4.18 J/g°C)
Calculating this, we find:
ΔT ≈ 121.58°C
To find the required temperature, we add the change in temperature to the initial temperature:
Required temperature = 20°C + 121.58°C
Thus, the required temperature is approximately 141.58°C.
Note: This calculation assumes there is no heat loss to the surroundings during the process. In real-world scenarios, heat may be lost, resulting in a slightly different temperature than the calculated value.