If the temperature of a 500.0-g sample of liquid water is raised 2.00c how much heat is absorbed by the water?The specific heat of liquid water is 4.184j/(g*c)
q= c • m • delta T
q= (4.184 J/g*C)(500.0g)(2.00*C)
q= 4184 J
q = mass H2O x specific heat H2O x delta T
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what the answer be doe
Well, well, well! Looks like we have a hot question here. Let me calculate that for you. So, you've got a 500.0-g sample of liquid water and you're raising its temperature by 2.00°C, right?
Alright, let's crunch some numbers. The specific heat of liquid water is 4.184 J/(g·°C), which means it takes 4.184 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
So, to find how much heat is absorbed by the water, we can use this formula:
Q = m · c · ΔT
Where:
Q is the heat absorbed (in joules),
m is the mass of the water (in grams),
c is the specific heat of water (in joules/gram·°C),
ΔT is the change in temperature (in °C).
Plugging in the numbers...
Q = 500.0 g · 4.184 J/(g·°C) · 2.00 °C
Ta-da! After some number crunching, we find that the amount of heat absorbed by the water is approximately 4,184 joules.
Water, you're on fire! But don't worry, it's just a temporary heatwave. Stay cool!
To find the heat absorbed by the water, you can use the formula:
Q = m * c * ΔT
Where:
Q is the heat absorbed by the water,
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature.
Given values:
m = 500.0 g (mass of water)
c = 4.184 J/(g*C) (specific heat capacity of water)
ΔT = 2.00 C (change in temperature)
Substituting the values into the formula, we get:
Q = 500.0 g * 4.184 J/(g*C) * 2.00 C
Multiplying the values, we have:
Q = 4,184 J/C * 2.00 C
Simplifying further, we find:
Q = 8,368 J
Therefore, the heat absorbed by the water is 8,368 Joules.