the amount of heat when 65g of water at 100c until it cools to 37c
q = heat released = mass H2O x specific heat H2O x (Tfinal-Tinitial)
To calculate the amount of heat transferred when 65g of water cools from 100°C to 37°C, you can use the formula:
Q = m * c * ΔT
Where:
Q is the heat transferred (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in joules per gram per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).
In this case, the substance is water. The specific heat capacity of water is approximately 4.18 J/g°C.
So, plugging in the values into the formula:
Q = 65g * 4.18 J/g°C * (37°C - 100°C)
First, let's calculate the difference in temperature:
ΔT = 37 - 100
ΔT = -63°C
Notice that the change in temperature is negative because the water is cooling down.
Now, substitute the values back into the formula:
Q = 65g * 4.18 J/g°C * (-63°C)
To get the heat transferred, multiply all the values together:
Q = -165,288 J
Therefore, the amount of heat transferred when 65g of water cools from 100°C to 37°C is approximately -165,288 joules (J). The negative sign indicates that heat is being lost during the cooling process.