Calculate how much heat 32.0 g of water absorbs when it is heated from 26.0°C to 73.5°C.
In joules, and then we have to calculate calories too, how is that done? I was absent. Thanks so much!
q = mass x specific heat x (Tfinal-Tinitial).
For joules, use 4.184 J/gram*C for specific heat. For calories, use 1.00 cal/gram*C
What's the constant for water though? Every time I look it up it's a different number
Oh nevermind you put it there. THANK YOU SO MUCH! You're a huge help!!!
I gave you the two constants. If you want q in joules, use specific heat = 4.184 J/g*C and if you want q in calories, use specific heat = 1.00 calorie/g*C.
You can look up specific heat in J/g, J/mol, cal/g, cal/mol, kg/mol, kg/g, etc. I gave you the two constants you need.
Thank you so much!!! You're really a lifesaver!
Is this is food calories?
no. Years ago the food industry got it all wrong. They talked about Big calories (identified by a CAPITAL letter C) and a small calorie (identified by a lower case c). But they made no real distinction and soon everyone (perhaps I should say the non-professionals) just called them "calories". A food calorie (the Big Calorie) actually is a kilocalorie. So when I drink a glass of no-fat milk, the label on the container says it is 80 calories. That is 80 kilocalories. So if we are to eat about 2000 calories a day to maintain our weight, that actually is 2000 kcal or 2,000,000 calories.
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To calculate the amount of heat absorbed by a substance, we can use the formula:
q = mcΔT
Where:
q is the heat absorbed (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in joules per gram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
For water, the specific heat capacity is approximately 4.18 joules per gram per degree Celsius.
To calculate the heat absorbed by the water, we will use the given values:
m = 32.0 g (mass of water)
ΔT = 73.5°C - 26.0°C = 47.5°C (change in temperature)
c = 4.18 J/g°C (specific heat capacity of water)
First, we calculate the heat absorbed in joules:
q = mcΔT
q = (32.0 g)(4.18 J/g°C)(47.5°C)
q ≈ 6388.4 J
So, the amount of heat absorbed by 32.0 g of water when heated from 26.0°C to 73.5°C is approximately 6388.4 joules.
To convert joules to calories, we use the conversion factor that 1 calorie (cal) is equal to 4.184 joules (J).
To calculate the heat absorbed in calories, we can use the formula:
q (calories) = q (joules) / conversion factor
q (calories) ≈ 6388.4 J / 4.184 J/cal
q (calories) ≈ 1528.96 cal
Therefore, the amount of heat absorbed is approximately 6388.4 joules or 1528.96 calories.