If the warm water in this experiment (Im not quite sure if you need more information) is at 98C, how much heat must be relesed in order for the temperature of 100. mL of water to decrease to 32C ?
q = mass water x specific heat water x delta T.
100 mL H2O = 100 g.
specific heat water = 4.184 J/g*C
delta T = 98-32.
To find out how much heat must be released for the temperature of water to decrease from 98°C to 32°C, you can use the equation:
Q = m * c * ΔT
Where:
Q is the heat transferred (in joules),
m is the mass of the water (in grams),
c is the specific heat capacity of water (approximately 4.18 J/g°C),
ΔT is the change in temperature (in °C).
First, you need to convert the volume of water (100 mL) to its mass. Since the density of water is 1 g/mL, the mass can be calculated as:
Mass (m) = Volume (V) * Density (D)
Mass (m) = 100 mL * 1 g/mL = 100 grams
Now you have the mass of water (m) = 100 g, the specific heat capacity of water (c) = 4.18 J/g°C, and the change in temperature (ΔT) = 98°C - 32°C = 66°C.
Plugging these values into the equation:
Q = 100 g * 4.18 J/g°C * 66°C
Q = 275928 J
Therefore, approximately 275,928 joules of heat must be released in order for the temperature of 100 mL of water to decrease from 98°C to 32°C.