I raised the temperature of 150mL of water from 25.000*C to 27.3*C How many Calories or Joules is it

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Sita

I raised the temperature of 150mL of water from 25.000C to 27.3C How many Calories or Joules is it

I raised the temperature of 150mL of water from 25.000*C to 27.3*C How many Calories or Joules of heat did i apply to get the temperature of water to change If there is formulas please tell me it or walk me through what you did to get the answer, if not the answer will work. The reason i ask this is am stuck in a lab were we have to determine what a unknown sample of metal is and i have to Use Specific heat to determine it I have every thing but “q” in the equation C= q / (m * (delta T)) I have tried to email my teacher but I don’t think I will reach her in time

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Sita:

thank you i now have what i need

Penny:

This link should help you, it describes what specific heat and it has a writen out equation, all you do is fill in the blanks which you should know how to do, if you wernt a slaker! JK
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html

Sita:

slacker ? try overachiever who is kicking himself in his rear end for not rembering this

Penny:

i said JK! god, people cant take a joke anymore, hope you try the link

Kevhav1:

345 calories
First to find out M(mass of water)
you need the formula
M=volume x density = vxd = 150 x 1 = 150 gm
Now, Specific heat of water(s)=1 calorie/gram °C
So, Heat applied(q)= MsΔt
= 150 x 1 x (27.3 - 25)
= 150 x 1 x 2.3 calories
= 345 calories
Now to find out the specific heat of metal(C)
use
C = q/mxΔt
= 345/mxΔt

Melly:

I only know the definition of a calorie: the amount of energy needed to raise 1 gram of water 1 degree centigrade.

Colder:

Total energy required to raise the temperature of water by T degrees centigrades = mass of water * specific heat of water * T = 150 grams * 1 calorie/(gm*deg centigrade)*2.3 = 345 calories. (notice how units cancel out)

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I raised the temperature of 150mL of water from 25.000C to 27.3C How many Calories or Joules is it

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