How much heat does it take to warm up a 300g glass of water by 30 degrees C?
(what is the formula for the specific heat of water?)
specific heat of water is a physical constant of nature, look it up.
thats not very helpful, will you at least answer the first question?
To calculate the heat required to warm up a substance, you can use the formula:
Q = m * c * ΔT
Where:
Q is the heat energy required (in joules),
m is the mass of the substance (in grams or kilograms),
c is the specific heat capacity of the substance (in joules per gram per degree Celsius or joules per kilogram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius or Kelvin).
In this case, we have:
m = 300g (mass of water)
c = specific heat of water
The specific heat of water is approximately 4.18 joules per gram per degree Celsius or 4.18 J/g°C.
ΔT = 30°C (change in temperature)
Now let's calculate the heat required using the formula:
Q = m * c * ΔT
Q = 300g * 4.18 J/g°C * 30°C
By simplifying the equation, we get:
Q ≈ 37,740 J
Therefore, it would take approximately 37,740 joules of heat to warm up a 300g glass of water by 30 degrees Celsius.
The specific heat capacity of water can be found in reference tables or by doing experiments. It is a characteristic property of the substance that determines how much heat energy it can absorb or release per unit mass and per degree temperature change.