Question
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
Answers
GPT 3.5
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.
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.