6 CHEM QUESTIONS ONLY!!! I DESPERATELY NEED A GENIUS FOR THIS?

there are 2 equations you need to know

.. (1) Q = m Cp dT... . for substances warming or cooling WITHOUT a phase change

.. (2) Q = m dHvap.... (or dHfusion.. or dHsublimation.. etc) for substances undergoing

... .. .. . ... ... ... ... .. .a phase change at constant temperature

*** 1 ***

.. Q = 5.0g x 4.184 J/g°C x (75°C - 25°C) = ___ J

*** 2 ***

.. Q = 5.0g x 1 cal/g°C x (75°C - 25°C) = ___ cal

*** 3 ***

.. Q = 35g x 333 J/g = ___ J

*** 4 ***

.. Q = 85g x 539.4 cal/g = ___ cal

*** 5 ***

.. Q = 25g x 4.184 J/g°C x (60°C - 10°C) = ___ J

*** 6 ***

.. Q = 50g x 79.72 cal/g = ___ cal

According to a table in my physics book, the specific heat of water is 4.186 J/(g * ˚C). To solve this type of problem, use the following equation.

Q = mass * specific heat * ∆ T

∆ T = 75 – 25 = 50˚

Q = 5 * 4.186 * 50 = 1,046.5 joules

2. How many calories are given off by the water in problem 1?(specific heat of water =1.0 cal/ g degrees celcius) Use the same equation.

Q = 5 * 1 * 50 = 250 calories

One joule is equal to 4.186 calories. You can use this to convert joules to calories. If you multiply 250 by 4.186, you will get the answer to the first problem. I hope this will be helpful for you.

3. how many joules does it take to melt 35 g of ice at 0 degrees celcius? (heat of fusion=333 j/ g)

Q = mass * heat of fusion = 35 * 333 = 11,655 J

4. how many calories are given off when 85 g of steam condense to liquid water? (heat of vaporization=539.4 cal/ g)

Q = mass * heat of vaporization = 85 * 539.4

5. How many joules of heat are necessary to raise the temperature of 25 g of water from 10 degrees celcius to 60 degrees celcius?

Use the equation for problem #1.

6. How many calories are given off when 50 g of water at 0 degrees freezes? (heat of fusion =79.72 cal/ g)

Since the water is at its freezing point, you can use the same equation as for problem #3

Q = 50 * 79.72