When 18 grams of copper absorbs 1 kJ (1000 J) of energy, the temperature increases from 35.0°C to 179.3 °C. What is the specific heat of copper?

Question

Avery Adams · Accepted Answer

#"0.39 J g"^(-1)""^@"C"^(-1)#

The specific heat of a substance tells you the amount of heat needed to increase the temperature of #"1 g"# of said substance by #1^@"C"#.

This implies that your ultimate goal here is to figure how much heat is needed to increase the temperature of #"1 g"# of copper by #1^@"C"#.

Now, you know that #"1000 J"# of heat increased the temperature of #"18 g"# of copper by

#179.3^@"C" - 35.0^@"C" = 144.3^@"C"#

so a good place to start would be to figure out the amount of heat needed to increase the temperature of #"1 g"# of copper by #144.3^@"C"#.

#1 color(red)(cancel(color(black)("g"))) * overbrace("1000 J"/(18color(red)(cancel(color(black)("g")))))^(color(blue)("to increase the temperature by 144.3"^@"C")) = "55.556 J"#

So, you know that if you add #"55.556 J"# of heat to #"1 g"# of copper, you will increase its temperature by #144.3^@"C"#.

This implies that in order to increase the temperature of #"1 g"# of copper by #1^@"C"#, you will need

#1 color(red)(cancel(color(black)(""^@"C"))) * overbrace("55.556 J"/(144.3color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 1 g of copper")) = "0.385 J"#

Therefore, you can say that copper has a specific heat of

#c_ "copper" = "0.39 J g"^(-1)""^@"C"^(-1)#

This tells you that you need #"0.39 J"# of heat to increase the temperature of #"1 g"# of copper by #1^@"C"#

I'll leave the answer rounded to two sig figs, the number of sig figs you have for the mass of copper, but keep in mind that you only have one significant figure for the energy absorbed by the sample.

This is an excellent result because the specific heat of copper is listed as being equal to #"0.39 J g"^(-1)""^@"C"^(-1)#.

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Luna Chen · Answer

undefined To find the specific heat of copper, use the formula:

$$ q = mc\Delta T $$

Where:
- $ q $ = heat absorbed (in joules)
- $ m $ = mass (in grams)
- $ c $ = specific heat (in J/g°C)
- $ \Delta T $ = change in temperature (in °C)

Rearranging the formula to solve for $ c $:

$$ c = \frac{q}{m\Delta T} $$

Given:
- $ q = 1000 \, J $
- $ m = 18 \, g $
- $ \Delta T = 179.3°C - 35.0°C = 144.3°C $

Plug in the values:

$$ c = \frac{1000 \, J}{18 \, g \times 144.3°C} $$

$$ c \approx \frac{1000 \, J}{2590.4 \, J} $$

$$ c \approx 0.386 \, J/g°C $$

So, the specific heat of copper is approximately $ 0.386 \, J/g°C $.