# Question #8f7d4

##### 1 Answer

#### Answer:

#### Explanation:

**QUICK ANSWER**

If you take

#135 color(red)(cancel(color(black)("g"))) * (T_f - 21)color(red)(cancel(color(black)(""^@"C"))) = 82.5 color(red)(cancel(color(black)("g"))) * (89 - T_f)color(red)(cancel(color(black)(""^@"C")))#

This will get you

#T_f = (82.5 * 89 + 135 * 21)/(135 + 82.5)#

#color(darkgreen)(ul(color(black)(T_f = 47))) " " -># rounded to twosig figs

You can thus say that the final temperature of the sample will be

**THE DETAILED VERSION**

Assuming that no heat is lost to the surroundings, you can say that the heat **lost** by the warmer sample will be **equal** to the heat **gained** by the colder sample.

#color(blue)(ul(color(black)(q_"gained" = - q_"lost")))" "color(darkorange)("(*)")# The minus sign is used here because

heat lostcarries a negative sign.

Now, the amount of heat gained/lost depends on the **mass** of the substance, **specific heat**, **change in temperature**, *initial temperature* from the *final temperature*.

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

If you take **final temperature** of the sample, you can say that the **heat gained** by the colder sample will be

#q_"gained" = "135 g" * c_"water" * (T_f - 21)^@"C"#

Similarly, the **heat lost** by the warmer sample will be

#q_"lost" = "82.5 g" * c_"water" * (T_f - 89)^@"C"#

Use equation **do not** forget about the minus sign!

#135 color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(c_"water"))) * (T_f - 21) color(red)(cancel(color(black)(""^@"C"))) = - 82.5 color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(c_"water"))) * (T_f - 89)color(red)(cancel(color(black)(""^@"C")))#

This simplifies to

#135 * (T_f - 21) = - 82.5 * (T_f - 89)#

which is equivalent to

#135 * (T_f - 21) = 82.5 * (89 - T_f)#

Once again, you will find that the final temperature of the sample will be

The answer must be rounded to two sig figs, the number of sig figs you have for the two temperatures.