q = mass Mg x specific heat Mg x delta T
2850 = 55.4 x 1.020 x delta T
Solve for delta T.
If 2850 J of heat energy was applied to 55.4g of magnesium, how much
would the temperature increase?
(cMg = 1.020 J/goC)
5 answers
so i just multiply those?
55.4 x 1.020= 56.508
55.4 x 1.020= 56.508
2850 = 55.4 x 1.020 x delta T
well, that gets part of it would it doesn't solve for delta T.
2850 = 55.4 x 1.020 x delta T
2850 = 56.508 x delta T
and you solve for delta T.
well, that gets part of it would it doesn't solve for delta T.
2850 = 55.4 x 1.020 x delta T
2850 = 56.508 x delta T
and you solve for delta T.
I'm new to this material and the course, what is delta x? how do I solve for it?
There is no delta x there? that x is a times sign. New to course or not this is just algebra. Suppose you have an equation that is
4y = 28 and you want to solve for y, the unknown. The easy way to that is
y = 28/4 = 7 but if you don't know how to do that you do it this way.
4y = 28. Doing anything to BOTH sides of the equation doesn't change anything so divide both side of the equation by 4 like this.
4y/4 = 28/4 and
y = 7. Do that with this equation.
2850 = 56.508 x delta T. The unknown is delta T
2850/56.508 = 56.609 x delta T/56.508
2850/56.508 = delta T
Divide to find delta T which is the change in temperature. That's what the problem asked you to calculate.
4y = 28 and you want to solve for y, the unknown. The easy way to that is
y = 28/4 = 7 but if you don't know how to do that you do it this way.
4y = 28. Doing anything to BOTH sides of the equation doesn't change anything so divide both side of the equation by 4 like this.
4y/4 = 28/4 and
y = 7. Do that with this equation.
2850 = 56.508 x delta T. The unknown is delta T
2850/56.508 = 56.609 x delta T/56.508
2850/56.508 = delta T
Divide to find delta T which is the change in temperature. That's what the problem asked you to calculate.