The amount of heat energy lost by the water as it cools from 70°C to 10°C can be calculated using the formula:
Q = mcΔT
where Q is the heat energy, m is the mass of the water (in grams), c is the specific heat capacity of water (4.2 J/g°C), and ΔT is the change in temperature (in °C).
ΔT = Final temperature - Initial temperature
Substituting the values gives:
Q = (2 g) * (4.2 J/g°C) * (10°C - 70°C)
Q = (2 * 4.2 * -60) J
Q = -504 J
So the amount of heat energy lost when 2 gram of water is cooled from 70°C to 10°C is -504 J.
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Answer:
The amount of heat energy lost by the water as it cools from 70°C to 10°C can be calculated using the formula:
Q = mcΔT
where Q is the heat energy, m is the mass of the water (in grams), c is the specific heat capacity of water (4.2 J/g°C), and ΔT is the change in temperature (in °C).
ΔT = Final temperature - Initial temperature
Substituting the values gives:
Q = (2 g) * (4.2 J/g°C) * (10°C - 70°C)
Q = (2 * 4.2 * -60) J
Q = -504 J
So the amount of heat energy lost when 2 gram of water is cooled from 70°C to 10°C is -504 J.