A similar question to that in problem 8, and we are going to use the following equation:

 q = m × C_s × ΔT

Where, q is the amount of heat in J, m is the mass, C_s is the specific heat capacity of the material in J/g ^oC , and ΔT is the change in temperature (T_final – T_initial). Remember, the specific heat is the quantity of heat energy necessary to increase the temperature of 1 g of a substance by 1°C.

All we need to do is to plug the numbers into the equation of the heat:

\[{\rm{q}}\;{\rm{ = }}\;{\rm{540}}{\rm{.6}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{0}}{\rm{.450}}\;{\rm{J}}}}{{\cancel{{\rm{g}}}\;\cancel{{^{\rm{o}}{\rm{C}}}}}}\;{\rm{ \times }}\;{\rm{(84}}{\rm{.3}}\;{\rm{ – }}\;{\rm{20}}{\rm{.0)}}\;\cancel{{^{\rm{o}}{\rm{C}}}}\;{\rm{ = }}\;{\rm{15}}{\rm{.6}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{{\rm{3}}\;}}{\rm{J}}\]

To convert the J to kJ, we need to divide it by 1,000, so it is 15.6 kJ

We are going to rearrange the following equation for the relation of heat, specific heat capacity, and temperature change:

 q = m × C × ΔT

\[{\rm{C}}\;{\rm{ = }}\;\frac{{\rm{q}}}{{{\rm{m}}\;{\rm{\Delta T}}}}\]

\[{\rm{C}}\;{\rm{ = }}\;\frac{{{\rm{481}}\;{\rm{J}}}}{{{\rm{17}}{\rm{.0}}\;{\rm{g}}\;{\rm{(67}}{\rm{.0}}\;{\rm{ – }}\;{\rm{25}}{\rm{.0)}}{\;^{\rm{o}}}{\rm{C}}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.674}}\;{\rm{J}}}}{{{\rm{g}}{\;^{\rm{o}}}{\rm{C}}}}\;\]

Because the combustion of butane increased the temperature of a bomb calorimeter, it is an exothermic reaction. This means, the heat goes from the reaction to the calorimeter and assuming there is no heat loss, the heat released by the reaction is equal to the heat absorbed by the calorimeter:

-q_reaction  = q_cal

Since the reaction occurs under conditions of constant volume, q_rxn = ΔE_rxn, so we need to calculate the q_cal using the following formula:

q_cal = C_cal x ΔT

q_cal = 2.36 kJ/ °C x 7.73 °C = 18.2 kJ

Therefore, q_rxn = -18.2 kJ

This is the change in the internal energy of the reaction for that specific amount of butane that was burned.

To get ΔE_rxn per mole of butane, we need to divide q_rxn by the number of moles that actually reacted. To find the number of moles, we use the mas and molar mass of butane:

n(C₄H₁₀) = 0.367 g/58.0 g/mol = 0.00633 mol

And after this, we divide the heat of the reaction by the number of moles of butane, to get the heat of the reaction per mole of butane:

ΔE_rxn = 18.2 kJ/0.00633 mol = 2.88 x 10³ kJ/mol

Alternatively, we can set up a cross-multiplication correlation:

0.00633 mol C₄H₁₀ – 18.2 kJ

1 mol C₄H₁₀ – X kJ

X = 2.88 x 10³ kJ/mol

Since the ice/water is already at 0 ^oC (melting/freezing temperature), the added heat does not change the temperature of the ice and water mixture, but rather it is used for the transition from solid to liquid.

To calculate the amount of energy required to achieve a state change, in this case melting, of the substance that is already at the state-change temperature, we use the following formula:

q = mΔH_fus

where q is the quantity of heat energy, m is the mass, and ΔH_fus is the enthalpy of fusion, sometimes called the heat of fusion. It is the energy required to fuse (melt) a gram (or a mole) of a substance.
Plugging the numbers in the formula, we get:

q = 40.0 g x 334.0 J/g = 13,360 J

Rounding off to three significant figures, we have q = 1.34 x 10³ J

We mentioned in the previous problem, that since the ice/water is already at 0 ^oC (melting/freezing temperature), the added heat does not change the temperature of the ice and water mixture, but rather it is used for the transition from solid to liquid. This is not the case here because the ice is at -15.0 oC and before melting, it needs to be first warmed to 0 ^oC. Therefore, there are two stages in this process; 1) heating the ice to 0 ^oC, 2) melting the ice.

The overall heat of this process then is:

q = q_heating + q_melting

q_heating is calculated by the formula we have already used several times:

q_heating = m × C × ΔT

\[{{\rm{q}}_{{\rm{heating}}}}\;{\rm{ = }}\;{\rm{36}}{\rm{.0}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{2.10\;{\rm{J}}}}{{\cancel{{\rm{g}}}\;\cancel{{^{\rm{o}}{\rm{C}}}}}}\;{\rm{ \times }}\;{\rm{(0}}\;{\rm{ – }}\;( – {\rm{15}}{\rm{.0)}}\;\cancel{{^{\rm{o}}{\rm{C}}}}\;{\rm{ = }}\;{\rm{1,134}}\;{\rm{J}}\]

To calculate the amount of energy required to melt the ice, we use the following formula:

q = mΔH_fus

\[{{\rm{q}}_{{\rm{melting}}}}\;{\rm{ = }}\;{\rm{36}}{\rm{.0}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{334.0\;{\rm{J}}}}{{\cancel{{\rm{g}}}\;}}\;{\rm{ = }}\;{\rm{12,024}}\;{\rm{J}}\]

q = 1134 J + 12,024 J = 13,158 J

Rounding of to three significant figures, we have q = 1.32 x 10⁴ J or 13.2 kJ