# A 40 g block of ice is cooled to -76°C. and is then added to 640 g of water in an 80 g copper calorimeter at a temperature of 28°C. What is the final temperature of the system consisting of the ice, water, and calorimeter?

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Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g ·°C = 2090 J/kg°C. specific heat of copper at 25 degree celcius is 0.9201731 calorie/g C

Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g ·°C = 2090 J/kg°C. specific heat of copper at 25 degree celcius is 0.9201731 calorie/g C

##### 1 Answer

#### Explanation:

**!! LONG ANSWER !!**

The idea here is that the heat **absorbed** by the ice will be **equal** to the heat **given off** by the liquid water and the calorimeter.

#color(blue)(q_"gained" = - q_"given off")#

This is equivalent to

#color(blue)(q_"ice" = -(q_"water" + q_"copper")" " " "color(orange)("(*)")#

The *minus sign* is used here because heat **given off** carries a minus sign.

Now, the trick is to realize that your sample of ice will go from

solidat#-76^@"C"# tosolidat#0^@"C"# solidat#0^@"C"# toliquidat#0^@"C"# liquidat#0^@"C"# toliquidat a final temperature#T_f#

The sample of liquid water will go from

liquidat#28^@"C"# toliquidat a final temperature#T_f#

Similarly, the calorimeter will go from

solidat#28^@"C"# tosolidat a final temperature#T_f#

Now, the equation that you can use looks like this

#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**

Let's start with the heat **absorbed** by the ice. In order to heat the ice from solid at

#q_"ice 1" = m_"ice" * c_"ice" * DeltaT_"ice 1"#

Convert the *specific heat* of ice from *joules per kilogram Celsius* to *joules per gram Celsius* first

#2090 "J"/(color(red)(cancel(color(black)("kg"))) ""^@"C") * (1color(red)(cancel(color(black)("kg"))))/(10^3"g") = "2.090 J g"^(-1)""^@"C"^(-1)#

Plug in your values to find

#q_"ice 1" = 40 color(red)(cancel(color(black)("g"))) * "2.090 J" color(red)(cancel(color(black)("g"^(-1))))color(red)(cancel(color(black)(""^@"C"^(-1)))) * [0 - (-76)]color(red)(cancel(color(black)(""^@"C")))#

#q_"ice 1" = "6353.6 J"#

In order to calculate the heat absorbed when you go from *solid* ice at *liquid* water at **enthalpy of fusion**,

#DeltaH_"fus" = "333.55 J g"^(-1)#

https://en.wikipedia.org/wiki/Enthalpy_of_fusion

You can thus say that the heat needed to convert the solid ice to liquid water at its melting point,

#40 color(red)(cancel(color(black)("g"))) * "333.55 J"/(1color(red)(cancel(color(black)("g")))) = "13,342 J"#

Next, calculate the heat needed to warm the water from liquid at **specific heat**

#c_"water" = "4.184 J g"^(-1)""^@"C"^(-1)#

You will have

#q_"ice 3" = 40 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 0)color(red)(cancel(color(black)(""^@"C")))#

#q_"ice 3" = (167.4 * T_f)" J"#

Now focus on the heat **given off** by the water that was initially present in the calorimeter. You will have

#q_"water" = m_"water" * c_"water" * DeltaT_"water"#

Plug in your values to get

#q_"water" = 640 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 28)color(red)(cancel(color(black)(""^@"C")))#

#q_"water" = 2677.8 * (T_f - 28)" J"#

Finally, calculate the heat **given off** by the calorimeter itself. Make sure that you convert the specific heat of copper from *calories per gram Celsius* to *joules per gram Celsius*

#0.09201731 color(red)(cancel(color(black)("cal")))/("g" ""^@"C") * "4.184 J"/(1color(red)(cancel(color(black)("cal")))) = "0.385 J g"^(-1)""^@"C"^(-1)#

The heat given off by the calorimeter will thus be

#q_"copper" = m_"copper" * c_"copper" * DeltaT_"copper"#

Plug in your values to get

#q_"copper" = 80 color(red)(cancel(color(black)("g"))) * "0.385 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (T_f - 28) color(red)(cancel(color(black)(""^@"C")))#

#q_"copper" = 30.8 * (T_f - 28)" J"#

Now it's time to put all this together using equation

#overbrace((q_"ice 1" + q_"ice 2" + q_"ice 3"))^(color(blue)("total heat absorbed by the ice")) = "6353.6 J" + "13,342 J" + (164.7 * T_f)" J"#

#q_"ice total" = (19696 + 164.7 * T_f)" J"#

This will get you

#(19696 + 164.7 * T_f) color(red)(cancel(color(black)("J"))) = -[2677.8 * (T_f - 28) + 30.8 * (T_f - 28)]color(red)(cancel(color(black)("J")))#

All you have to do now is solve for

#19696 + 164.7 * T_f = -2677.8 * T_ f + 74978.4 - 30.8 * T_f + 862.4#

#2873.3 * T_f = 56144.8 implies T_f = 55144.8/2873.3 = 19.5402#

Therefore, you can say that the final temperature of the system will be

#color(green)(|bar(ul(color(white)(a/a)color(black)(T_f = 19.5^@"C")color(white)(a/a)|)))#

I'll leave the answer rounded to three **sig figs**, but keep in mind that your values only justify one sig fig for the final temperature..