# Question #11e6e

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that you can assume that the heat **lost** by the iron bar as it cools will be **equal** to the heat **gained** by the water as it warms.

#color(blue)(ul(color(black)(-q_"iron" = q_"water")))# The

minus signis used here becauseheat lostis, by definition, negative.

Your tool of choice here will be the equation

#color(blue)(ul(color(black)(q = m * c * DeltaT)))#

Here

#q# is the heat lost or gained by the substance#m# is themassof the sample#c# is thespecific heatof the substance#DeltaT# is thechange in temperature, defined as the difference between thefinal temperatureand theinitial temperatureof the sample

You should know that the specific heat of water is

#c_"water" = "1 cal g"^(-1)""^@"C"^(-1)#

If you take **final temperature** reached by the iron bar + water system, you can say that you have

For the iron bar

#q_"iron" = 45.0 color(red)(cancel(color(black)("g"))) * "0.11 cal" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 100)color(red)(cancel(color(black)(""^@"C")))#

#q_"iron" = 4.95 * (T_f - 100)" cal"#

For the water

#q_"water" = 90.0 color(red)(cancel(color(black)("g"))) * "1 cal" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 40)color(red)(cancel(color(black)(""^@"C")))#

#q_"water" = 90.0 * (T_f - 40)" cal"#

Since the heat lost by the iron bar must be equal to heat gained by the water, you will have -- **do not** forget about the minus sign used for the heat lost by the iron bar!

#-4.95 * (T_f - 100)color(red)(cancel(color(black)("cal"))) = 90.0 * (T_f - 40)color(red)(cancel(color(black)("cal")))#

This will get you

#-4.95 * T_f + 495 = 90.0 * T_f - 3600#

#94.95 * T_f = 4095 implies T_f = 4095/94.95 = 43.13#

Therefore, you can say that the iron bar will cool to a final temperature of

#color(darkgreen)(ul(color(black)(T_f = 43^@"C")))#

I'll leave the answer rounded to two **sig figs**, but keep in mind that you only have one significant figure for the starting temperatures of the iron bar and of the water.