# Question #cb561

##### 1 Answer

#### Explanation:

The idea here is that the heat **given off** by the metal as it cooled from **equal** to the heat **absorbed** by the water as it warmed from

#color(blue)(ul(color(black)(-q_"metal" = q_"water")))" " " "color(red)("(*)")#

The.minus signis used here because, by definition, heatgiven offcarries a negative sign

So, you should know that water has a specific heat of about

#c_"water"= "4.18 j g"^(-1)""^@"C"^(-1)#

You can use the information provided by the problem to figure out *how much heat* was absorbed by the water.

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

In your case, you know that the change in temperature for the water is equal to

#DeltaT = 36.5^@"C" - 29^@"C" = 7.5^@"C"#

which means that you have

#q_"water" = 41 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 7.5color(red)(cancel(color(black)(""^@"C")))#

#q_"water" = "1285.35 J"#

Now, according to equation

#-q_"metal" = "1285.35 J" implies q_"metal" = - "1285.35 J"#

Use the same equation to find the specific heat of the metal,

#c_"metal" = (q_"metal")/(m_"metal" * DeltaT_"metal")#

In this case, you have

#DeltaT_"metal" = 36.5^@"C" - 84^@"C" = -46.5^@"C"#

This means that the metal has a specific heat equal to

#c_"metal" = (color(red)(cancel(color(black)(-)))"1285.35 J")/(37color(red)(cancel(color(black)("g"))) * (color(red)(cancel(color(black)(-)))46.5^@"C")) = color(darkgreen)(ul(color(black)("0.75 J g"^(-1)""^@"C"^(-1))))#

The answer is rounded to two **sig figs**.