# Question #087c4

##### 1 Answer

#### Explanation:

For starters, you need to make a note that the **specific heat** of water is equal to

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

This means that when the temperature of *liquid water* decreases by **given off**.

Now, your tool of choice here will be this equation

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

Here

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

You need to solve for

#DeltaT = q/(m * c)#

Plug in your values to find

#DeltaT = (1794 color(red)(cancel(color(black)("cal"))))/(250 color(red)(cancel(color(black)("g"))) * 1 color(red)(cancel(color(black)("cal"))) color(red)(cancel(color(black)("g"^(-1)))) ""^@"C"^(-1)) = 7.2^@"C"#

Now, you know that the sample is **losing heat**, so you can say that the final temperature will be **lower** than the initial temperature

#T_"final" = T_"initial" - DeltaT#

which, in your case, will be equal to

#T_"final" = 98.8^@"C" - 7.2^@"C" = color(darkgreen)(ul(color(black)(91.6^@"C")))#

The answer is rounded to one *decimal place*.

**SIDE NOTE** *It's important to notice that if you use*

#DeltaT = T_"final" - T_"initial"#

*you get*

#T_"final" = DeltaT + T_"initial"#

#color(red)(cancel(color(black)(T_"final" = 7.2^@"C" + 98.8^@"C" = 106^@"C"))) -># not good

*The problem here is that the sample is actually giving off heat, so you should use*

#q = - "1794 cal"#

*in the equation. The minus sign is needed because the final temperature is lower than the initial temperature. This will get you*

#DeltaT = (-1794 color(red)(cancel(color(black)("cal"))))/(250 color(red)(cancel(color(black)("g"))) * 1 color(red)(cancel(color(black)("cal"))) color(red)(cancel(color(black)("g"^(-1)))) ""^@"C"^(-1)) = -7.2^@"C"#

*which results in*

#T_"final" = - 7.2^@"C" + 98.8^@"C" = 91.6^@"C" -># good