If 200. grams of water is to be heated from 24.0°C to 100.0°C to make a cup of tea, how much heat must be added? The specific heat of water is 4.18 J/g∙°C
The amount of heat required can be calculated using the formula:
q = m * c * ΔT
where:
q is the heat in joules (J)
m is the mass of water in grams (200.0 g)
c is the specific heat of water (4.18 J/g∙°C)
ΔT is the change in temperature (100.0°C - 24.0°C = 76.0°C)
Substituting the given values into the formula:
q = 200.0 g * 4.18 J/g∙°C * 76.0°C
q = 63712 J
Therefore, approximately 63,712 J of heat must be added to heat the water from 24.0°C to 100.0°C.
To calculate the heat required to heat 200 grams of water from 24.0°C to 100.0°C, you can use the following formula:
Q = mcΔT
Where:
Q = heat (in joules)
m = mass of water (in grams)
c = specific heat of water (in J/g∙°C)
ΔT = change in temperature (in °C)
Let's plug in the given values:
m = 200 grams
c = 4.18 J/g∙°C
ΔT = (100.0°C - 24.0°C) = 76.0°C
Now, let's calculate the heat:
Q = (200 grams) x (4.18 J/g∙°C) x (76.0°C)
= 63344 J
Therefore, approximately 63344 joules of heat must be added to heat 200 grams of water from 24.0°C to 100.0°C.
To calculate the amount of heat needed to heat the water, you can use the formula:
Q = m * c * ΔT
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)
In this case, you have:
m = 200.0 grams
c = 4.18 J/g∙°C
ΔT = (final temperature) - (initial temperature) = 100.0°C - 24.0°C = 76.0°C
Now, substitute the values into the formula and calculate the result:
Q = (200.0 g) * (4.18 J/g∙°C) * (76.0°C)
Q = 62937.6 J
Therefore, to heat 200.0 grams of water from 24.0°C to 100.0°C, you need to add approximately 62937.6 Joules of heat energy.