# Question #1dcd8

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that the heat **lost** by the metal will be **equal to** the heat **absorbed** by the water.

#color(blue)(-q_"metal" = q_"water")#

Here the *minus sign* is used because, by definition, **heat lost** carries a negative sign.

Your tool of choice here will be the equation

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "# , where

*change in temperature*, defined as the difference between the **final temperature** and the **initial temperature**

You know the mass of water, its initial temperature, the final temperature of the system, and water's **specific heat**, which means that you can determine how much heat was needed to increase the temperature of the water by

#DeltaT_"water" = 16.19^@"C" - 14.00^@"C" = 2.19^@"C"#

Plug in your values to get

#q_"water" = 170.0color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 2.19color(red)(cancel(color(black)(""^@"C")))#

#q_"water" = "1556.214 J"#

So, this is how much heat was *absorbed* by the water, which can only mean that this is how much heat was **lost** by the metal.

You can thus say that

#-q_"metal" = "1556.214 J"#

The change in temperature for the metal will be

#DeltaT_"metal" = 16.19^@"C" - 79.00^@"C" = -62.81^@"C"#

This means that you have

#color(red)(cancel(color(black)(-)))"1556.214 J" = "170.0 g" * c_"metal" * (color(red)(cancel(color(black)(-)))62.81)^@"C"#

The specific heat of the metal will be

#c_"metal" = "1556.214 J"/("170.0 g" * 62.81^@"C")#

#c_"metal" = "0.14574 J g"^(-1)""^@"C"^(-1)#

Rounded to four **sig figs**, the answer will be

#c_"metal" = color(green)(|bar(ul(color(white)(a/a)"0.1457 J g"^(-1)""^@"C"^(-1)color(white)(a/a)|)))#