To determine the amount of boiling water needed to heat 1 metric ton (1,000 kilograms) of room temperature water from 20 degrees Celsius to 30 degrees Celsius, we can use the principle of heat transfer. The equation for heat transfer is:
Q = m * c * ΔT
Where: Q = heat transferred m = mass of the substance c = specific heat capacity of the substance ΔT = change in temperature
In this case, we want to find the heat transferred (Q) from the boiling water to the room temperature water. We assume that there is no heat loss during the process.
Let's assume the boiling water is at 100 degrees Celsius (boiling point) and the specific heat capacity of water is approximately 4.186 joules per gram per degree Celsius.
First, we need to convert the metric ton of water into kilograms: 1 metric ton = 1,000 kilograms
Now we can calculate the heat transfer required:
Q = m * c * ΔT Q = 1000 kg * 4.186 J/g°C * (30°C - 20°C)
Note that we need to convert the specific heat capacity from joules per gram per degree Celsius to joules per kilogram per degree Celsius:
1 g = 0.001 kg 4.186 J/g°C = 4.186 J/0.001 kg°C = 4186 J/kg°C
Substituting the values into the equation:
Q = 1000 kg * 4186 J/kg°C * 10°C Q = 41,860,000 J
Therefore, you would need to transfer approximately 41,860,000 joules of heat from the boiling water to the room temperature water to raise its temperature from 20°C to 30°C.
Please note that this calculation assumes ideal conditions with no heat loss or other factors that could affect the outcome. In reality, there may be some heat loss due to various factors, so this is an approximation.