A 150-L water heater is rated at 8 kW .If 20 percent of its heat escapes, how long does the heater take to raise the temperature of 150 L of water from 10C to 60C?
To solve this problem, we need to use the equation for heat transfer:
Q = mcΔT
where Q is the heat transferred, m is the mass of the substance (in this case, water), c is the specific heat capacity of water, and ΔT is the change in temperature.
Let's break down the problem step by step:
1. Calculate the mass of the water:
Given that the volume of the water is 150 L, we need to convert it to mass. The density of water is roughly 1 g/mL, or 1 kg/L.
So, the mass of the water is: mass = volume * density = 150 L * 1 kg/L = 150 kg.
2. Calculate the heat transferred:
The heat transferred (Q) is given by the equation Q = mcΔT.
Given that the specific heat capacity (c) of water is approximately 4.18 J/g·°C, or 4.18 kJ/kg·°C, we can substitute the values:
Q = (150 kg) * (4.18 kJ/kg·°C) * (60°C - 10°C) = 37.62 MJ.
3. Calculate the heat applied by the heater:
Since 20% of the heat escapes, the heater only applies 80% of the total heat needed.
So, the heat applied by the heater = 0.8 * Q = 0.8 * 37.62 MJ = 30.096 MJ.
4. Calculate the time taken by the heater to complete the heating process:
The power (P) of the heater is given as 8 kW.
The relationship between power, energy, and time is: Power = Energy / Time.
Rearranging the equation, we can solve for Time: Time = Energy / Power.
Using the heat applied by the heater (30.096 MJ) and converting it to energy in joules: 30.096 MJ * 1,000,000 J/MJ = 30,096,000 J.
Then, Time = 30,096,000 J / (8 kW * 1000 J/kW) = 3,762 seconds = 62.7 minutes.
Therefore, the heater takes approximately 62.7 minutes to raise the temperature of 150 L of water from 10°C to 60°C.