If 100.0 g of water at 90 deg. c is added to 50.0 g of water at 10 deg. c, estimate the final temperature of the water.
a point. a line would show progressing distance over passing time. since the object is not going anywhere it would just be a point. i would also assume that this point would be sitting at the 0 on the distance axis. (this assumes distance is the x axis)
if the x-axis is time and the y-axis is distance you will get a horizontal line because time will be passing but the distance will (y value) constant (at rest).
29.5x - 6162.55 = 16414.89 - 54.3x
83.8x = 22577.44
x = 269.4 k
in case, you're not sure what happened to the 2.53, i simply divided both sides by 2.53 first.
example #5: a sheet of nickel weighing 10.0 g and at a temperature of 18.0 °c is placed flat on a sheet of iron weighing 20.0 g and at a temperature of 55.6 °c. what is the final temperature of the combined metals? assume that no heat is lost to the surroundings.
this problem requires us to find the specific heats for nickel and iron. to do this, we will use this site. the values given are respectively, 0.54 j g¯1 °c¯1 and 0.46 j g¯1 °c¯1
notice that the units on the site are kj kg¯1 k¯1. in addition, notice that i wrote j g¯1 °c¯1. also, notice that there is no numerical difference when using either specific heat unit (the kj one or the j one). in other words:
one kj kg¯1 k¯1 = one j g¯1 °c¯1
the left-hand unit is the iupac-approved one; the one on the right-hand side is the one in most common use.
on to the solution:
qlost = qgain
(20.0) (55.6 - x) (0.46) = (10.0) (x - 18.0) (0.54)
9.2 (55.6 - x) = 5.4 (x - 18)
511.52 - 9.2x = 5.4x - 97.2
14.6x = 608.72
x = 41.7 °c
its d earths seasons
1. walls, doors