# How do I find the specific heat in this question?

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When a 57-gram piece of aluminum at 100oC is placed in water, it loses 735 calories of heat while cooling to 30oC. Calculate the specific heat of the aluminum

When a 57-gram piece of aluminum at 100oC is placed in water, it loses 735 calories of heat while cooling to 30oC. Calculate the specific heat of the aluminum

##### 1 Answer

#### Explanation:

The **specific heat** of aluminium tells you the amount of heat *needed or released* when the temperature of **changes** by

In your case, the temperature **decreases** by

#100^@"C" - 30^@"C" = 70^@"C"#

so the value of the specific heat will tell you the amount of heat given off when

So, the first thing to do here is to figure out the heat given off when **calories** were given off when the sample cooled by

#"735 cal"/(70^@"C") = "10.5 cal" ""^@"C"^(-1)#

This means that when the temperature of **calories** of heat are being given off.

Now all you have to do is figure out the amount of heat given off when the temperature of

#("10.5 cal" ""^@"C"^(-1))/"57 g" = "0.184 cal g"^(-1)""^@"C"^(-1)#

And there you have it -- aluminium has a specific heat of

#color(darkgreen)(ul(color(black)(c_"Al" = "0.184 cal g"^(-1)""^@"C"^(-1))))#

I'll leave the answer rounded to three **sig figs**, but keep in mind that you only have one significant figure for the initial temperature of the metal.

You can double-check your result by using 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

Now, notice that the change in temperature is actually equal to

#DeltaT = 30^@"C" - 100^@"C" = - 70^@"C"#

This is consistent with the fact that heat is being **lost** by the metal. However, you don't really need to use the minus sign when solving for the specific heat of a material.

That is the case because specific heat is meant to symbolize both heat *absorbed* when the temperature of **and** heat *given off* when the temperature of

In this regard, specific heat is not meant to indicate if heat is being lost or gained. Therefore, when calculating the specific heat of a material, you can go for

#c = q/(m * |DeltaT|)#

This will ensure that the specific heat comes out positive regardless if the material loses or gains heat.

So, plug in your values to find

#c = "735 cal"/("57 g" * |-70^@"C"|) = color(darkgreen)(ul(color(black)("0.184 cal g"^(-1)""^@"C"^(-1))))#