if you take a 1 liter bottle of water out of the refrigerator at just above 0 celsius., how much heat will it take to raise the water to 90 celsius, so you can make tea out of it.
Q = M*C*(T2 - T1)
= 1000 g * 1.00 cal/g C * 90 C
= 90,000 calories = 90 kilocalories
I am quite amazed. I appreciate your help!
To determine the amount of heat needed to raise the temperature of water from just above 0 degrees Celsius to 90 degrees Celsius, we can use the specific heat capacity formula:
Q = mcΔT
Where:
- Q is the amount of heat (in joules)
- m is the mass of water (in kilograms)
- c is the specific heat capacity of water (4.186 J/g°C or 4190 J/kg°C for liquid water)
- ΔT is the change in temperature (in degrees Celsius)
First, we need to convert the volume of water (1 liter) into its mass:
The density of water is approximately 1 gram/mL or 1000 kg/m³. Therefore, the mass of 1 liter (1000 mL) of water is 1000 grams or 1 kilogram.
Now that we have the mass (m = 1 kg), specific heat capacity (c = 4190 J/kg°C), and the change in temperature (ΔT = 90°C - 0°C = 90°C), we can calculate the amount of heat (Q) required:
Q = mcΔT
Q = 1 kg × 4190 J/kg°C × 90°C
Q = 376,100 J (or approximately 376.1 kJ)
So, it would take approximately 376,100 joules (or 376.1 kilojoules) of heat to raise the temperature of 1 liter of water from just above 0°C to 90°C.