How much heat is needed to bring 120 g of water from 33C to 45C
q=mc∆t=(120g)(4.186 J/g•C°)(45C-35C)
q=mc∆t=(120g)(4.186 J/g•C°)(45C-33C)
Typo 33 not 35
Well, it really depends on how hot your jokes are.
To calculate the amount of heat required to raise the temperature of a substance, you can use the formula:
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)
For water, the specific heat capacity is approximately 4.18 J/g°C.
Using the given values:
m = 120 g
ΔT = (45°C - 33°C) = 12°C
c (specific heat capacity of water) = 4.18 J/g°C
Plugging in the values into the formula:
Q = (120 g) x (4.18 J/g°C) x (12°C)
Q = 60,192 joules
Therefore, approximately 60,192 joules of heat is needed to bring 120 g of water from 33°C to 45°C.
To calculate the amount of heat needed to bring a substance from one temperature to another, we can use the equation:
Q = m * c * ΔT
Where:
Q is the heat (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)
In this case, we are dealing with water, which has a specific heat capacity of 4.18 J/g°C.
First, let's calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 45°C - 33°C
ΔT = 12°C
Now we can plug in the values into the equation:
Q = 120 g * 4.18 J/g°C * 12°C
Calculating this gives us:
Q = 6028.8 J
Therefore, it would take 6028.8 joules of heat to bring 120 grams of water from 33°C to 45°C.