A 1000-g iron frying pan is placed on a stove. The pan increases from 20°C to 130°C. If the specific heat of iron is 0.444 J/goC, how much heat was gained by the pan?

48,840 Joules

Explanation:

The equation you should use for this problem is Q=m c ΔT

Now just substitute based of of what you know.

We do not know what Q is. M equals 1,000 grams. C equals .444 the specific heat capacity. Delta T equals 110.

*To find T you must subtract 130-20 degrees to find the change.

The final equation should be: Q= 1000×.444×110

Now all you do is solve like you would normally and the answer you should get is: 48,840.

And so that how the answer comes out :D

i dont know its just needed points lol

Explanation:

I will put answer in th  repiles

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.

answer///carbon;