# 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?

## 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

Aug 21, 2016

${19.5}^{\circ} \text{C}$

#### Explanation:

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.

$\textcolor{b l u e}{{q}_{\text{gained" = - q_"given off}}}$

This is equivalent to

$\textcolor{b l u e}{{q}_{\text{ice" = -(q_"water" + q_"copper")" " " "color(orange)("(*)}}}$

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

• solid at $- {76}^{\circ} \text{C}$ to solid at ${0}^{\circ} \text{C}$
• solid at ${0}^{\circ} \text{C}$ to liquid at ${0}^{\circ} \text{C}$
• liquid at ${0}^{\circ} \text{C}$ to liquid at a final temperature ${T}_{f}$

The sample of liquid water will go from

• liquid at ${28}^{\circ} \text{C}$ to liquid at a final temperature ${T}_{f}$

Similarly, the calorimeter will go from

• solid at ${28}^{\circ} \text{C}$ to solid at a final temperature ${T}_{f}$

Now, the equation that you can use looks like this

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} q = m \cdot c \cdot \Delta T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$q$ - the amount of heat gained
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the 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 $- {76}^{\circ} \text{C}$ to solid at ${0}^{\circ} \text{C}$, you must provide it with

${q}_{\text{ice 1" = m_"ice" * c_"ice" * DeltaT_"ice 1}}$

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

$2090 {\text{J"/(color(red)(cancel(color(black)("kg"))) ""^@"C") * (1color(red)(cancel(color(black)("kg"))))/(10^3"g") = "2.090 J g"^(-1)""^@"C}}^{- 1}$

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")))

${q}_{\text{ice 1" = "6353.6 J}}$

In order to calculate the heat absorbed when you go from solid ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$, you must use water's enthalpy of fusion, $\Delta {H}_{\text{fus}}$, which is equal to

$\Delta {H}_{\text{fus" = "333.55 J g}}^{- 1}$

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, ${q}_{\text{ice 2}}$, is equal to

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 ${0}^{\circ} \text{C}$ to liquid at the final temperature ${T}_{f}$ by using water's specific heat

${c}_{\text{water" = "4.184 J g"^(-1)""^@"C}}^{- 1}$

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")))

${q}_{\text{ice 3" = (167.4 * T_f)" J}}$

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

${q}_{\text{water" = m_"water" * c_"water" * DeltaT_"water}}$

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")))

${q}_{\text{water" = 2677.8 * (T_f - 28)" J}}$

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

${q}_{\text{copper" = m_"copper" * c_"copper" * DeltaT_"copper}}$

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")))

${q}_{\text{copper" = 30.8 * (T_f - 28)" J}}$

Now it's time to put all this together using equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$. You will have

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"

${q}_{\text{ice total" = (19696 + 164.7 * T_f)" J}}$

This will get you

$\left(19696 + 164.7 \cdot {T}_{f}\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{J"))) = -[2677.8 * (T_f - 28) + 30.8 * (T_f - 28)]color(red)(cancel(color(black)("J}}}}$

All you have to do now is solve for ${T}_{f}$, the final temperature of the ice + water + calorimeter system

$19696 + 164.7 \cdot {T}_{f} = - 2677.8 \cdot {T}_{f} + 74978.4 - 30.8 \cdot {T}_{f} + 862.4$

$2873.3 \cdot {T}_{f} = 56144.8 \implies {T}_{f} = \frac{55144.8}{2873.3} = 19.5402$

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

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{T}_{f} = {19.5}^{\circ} \text{C}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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..