The water doesn’t go through any phase changes so, we can use the following formula for the relationship between the amount of heat needed to raise the temperature of a given substance and the corresponding temperature increase:

 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 and it is 4.18 J/g ^oC for water. Therefore, to find the amount of heat for a given mass, we multiply it by this mass.

Before using the numbers, make sure the units match. The C_s is given for a gram of water so, we need to convert the volume to mL first, and then to mass in g.

\[{\rm{V}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{L}}\;{\rm{ \times }}\;\frac{{{\rm{1,000}}\;{\rm{mL}}}}{{{\rm{1}}\;{\rm{L}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{mL}}\]

m = d x V

\[{\rm{m}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{mL}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{g}}}}{{{\rm{mL}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{g}}\]

And now, we can plug the numbers into the equation of the heat:

q = m × C_s × ΔT

\[{\rm{q}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{3}}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}{\rm{.18}}\;{\rm{J}}}}{{\cancel{{\rm{g}}}\;\cancel{{^{\rm{o}}{\rm{C}}}}}}\;{\rm{ \times }}\;{\rm{(100}}{\rm{.0}}\;{\rm{ – }}\;{\rm{25}}{\rm{.0)}}\;\cancel{{^{\rm{o}}{\rm{C}}}}\;{\rm{ = }}\;{\rm{4}}{\rm{.70}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{{\rm{5}}\;}}{\rm{J}}\]

To convert the J to kJ, we need to divide it by 1,000, so it is 4.70 x 10² kJ

The basis of solving this problem is the assumption that q_lost = q_gained. The heat flows from the warmer sample (q_lost) to the cooler one (q_gained), and the amount of heat lost by the warmer sample is equal to the amount of heat gained by the cooler sample. This means that there is no heat lost to the surroundings.

Recall that q = m · C · ΔT, where m is the mass, C is the specific heat, and ΔT is the temperature change.

We can also write this as q_lost + q_gained = 0 or

q₁ + q₂ = 0

For our problem, q₁ = mCΔT₁ , q₂ = mCΔT₂

so, mCΔT₁  + mCΔT₂ = 0

ΔT₁ = T_f – 90, and ΔT₂ = T_f – 25

Where T_f  is the final temperature of the system, so it is the same for the worm and cold samples.

Plug in the numbers and solve for T_f  in the following equation:

40 g x C x (T_f – 90) + 60 g x C x (T_f – 25) = 0

Because C (heat capacity of water) appears on both sides of the equation and can be canceled out.

40 g x (T_f – 90) + 60 g x (T_f – 25) = 0

T_f  = 51 °C

Notice that the equation can also be set up by using kelvin (K).

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

a) The ΔH value for a reaction is given specifically based on the coefficients in the balanced equation. Because the coefficient in front of H₂O is a 4, the formation of one mole of water will release:

-1452.8 kJ ÷ 4 = 363.2 kJ

b) According to the equation, when 3 moles of oxygen react, 8 kJ energy is released. Therefore, for the reaction of one-mole oxygen,

ΔH = -1452.8 kJ ÷ 3 = 484.27 kJ

The ΔH value for a reaction is given specifically based on the coefficients in the balanced equation. So, in this reaction, the combustion of two moles of SO₂ produces 198 kJ of heat. What we need to do here is first convert the mass of SO₂ to moles, and then calculate the amount of heat that will be released from this amount considering that burning two moles of SO₂ gives 198 kJ of heat.

n(SO₂) = 44.8 g/64.1 g/mol = 0.699 mol

Let’s now set up a cross multiplication, which would read as “2 mol SO₂ gives 198 kJ heat, 0.699 mol SO₂ will give an unknown amount of heat:

2 mol SO₂ – 198 kJ

0.699 mol SO₂ – X kJ

X = 69.2 kJ

Alternatively, we can do this by unit conversion method:

\[\Delta H\; = \;\frac{{{\rm{198}}\;{\rm{kJ}}}}{{\cancel{{{\rm{2}}\;{\rm{mol}}\;{\rm{S}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.699 }}\cancel{{{\rm{mol}}\;{\rm{S}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ = }}\;{\rm{69}}{\rm{.2}}\;{\rm{kJ}}\]

So, 69.2 kJ heat will be released if 44.8 g (0.699 mol ) of SO₂ is reacted according to the given chemical equation.

The ΔH value for a reaction is given specifically based on the coefficients in the balanced equation. So, in this reaction, we need to determine the combustion of two moles of C₆H_6.

Convert the mass of C₆H₆ to moles, and then calculate the amount of heat that will be released from 2 moles of C₆H₆ considering that burning 27.3 g C₆H₆ gives 1144 kJ of heat.

n(C₆H₆) = 27.3 g/78.1 g/mol = 0.350 mol

Let’s now set up a cross multiplication, which would read as “0.350 mol C₆H₆ gives 1144 kJ heat, 2 mol C₆H₆ will give an unknown amount of heat:

0.350 mol C₆H₆ – 1144 kJ

2 mol C₆H₆ – X kJ

X = 6.54 x 10³ kJ

Alternatively, we can do this by unit conversion method:

\[\Delta H\; = \;\frac{{{\rm{1144}}\;{\rm{kJ}}}}{{\cancel{{{\rm{0}}{\rm{.350}}\;{\rm{mol}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{6}}}}}}}\;{\rm{ \times }}\;{\rm{2}}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{6}}}}}\;{\rm{ = }}\;{\rm{6}}{\rm{.54}}\;{\rm{ \times }}\;{10^3}\;{\rm{kJ}}\]

The ΔH value for a reaction is given specifically based on the coefficients in the balanced equation which imply their mole numbers. Therefore, converting the mass to moles is always going to be a good idea.

n(HCl) = 233.6/36.5 g/mol = 6.40 mol

According to the chemical equation, for every three moles of HCl forming, there is 334 kJ heat released. So, we need to find how much heat will be released if 6.40 mol HCl is formed.

We can set up a cross multiplication, which would read as “3 mol HCl gives 334 kJ heat, 6.40 mol HCl will give an unknown amount of heat:

3 mol HCl – 334 kJ

6.40 mol HCl – X kJ

X = 713 kJ

Alternatively, we can do this by unit conversion method:

\[\Delta H\; = \;\frac{{{\rm{334}}\;{\rm{kJ}}}}{{{\rm{3 }}\cancel{{{\rm{mol HCl}}}}}}\;{\rm{ \times }}\;{\rm{6}}{\rm{.40}}\cancel{{{\rm{mol}}\;{\rm{HCl}}}}\;{\rm{ = }}\;713\;{\rm{kJ}}\]

First, let’s convert all the quantities to moles.

n(MgCO₃) = 12.65 g/84.3 g/mol = 0.150 mol

The moles of HCl are calculated using the equation for molarity:

M = n/V

n = MV

n(HCl) = MV = 0.400 M x 0.650 L = 0.260 mol

We have the moles of both reactants, and therefore, we need to find the limiting reactant in order to do the calculations for the heat of the reaction. Remember, the limiting reactant is the one that gives less product. So, to find the LR, we determine whether the MgCO₃ or HCl could produce less product based on their moles and the reaction stoichiometry. Let’s do the calculations based on the number of MgCl₂ moles formed:

\[{\rm{n}}\left( {{\rm{MgC}}{{\rm{l}}_{\rm{2}}}\;{\rm{from}}\;{\rm{MgC}}{{\rm{O}}_{\rm{3}}}} \right){\rm{\; = }}\;{\rm{0}}{\rm{.150}}\;\cancel{{{\rm{mol}}\;{\rm{MgC}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MgC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{MgC}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.150}}\;{\rm{mol}}\]

\[{\rm{n}}\left( {{\rm{MgC}}{{\rm{l}}_{\rm{2}}}\;{\rm{from}}\;{\rm{HCl}}} \right){\rm{\; = }}\;{\rm{0}}{\rm{.260}}\;\cancel{{{\rm{mol}}\;{\rm{HCl}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MgC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{HCl}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.130}}\;{\rm{mol}}\]

So, HCl gives less MgCl₂, therefore, it is the limiting reactant, and we need to calculate the amoiunt of heat beased on 0.260 mol HCl.

We can set up a cross multiplication, which would read as “2 mol HCl gives 112 kJ heat, 0.260 mol HCl will give an unknown amount of heat:

2 mol HCl – 112 kJ

0.260 mol HCl – X kJ

X = 14.6 kJ

Alternatively, we can do this by unit conversion method:

\[\Delta H\; = \;\frac{{{\rm{112}}\;{\rm{kJ}}}}{{{\rm{2 }}\cancel{{{\rm{mol}}\;{\rm{HCl}}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.260}}\cancel{{{\rm{mol}}\;{\rm{HCl}}}}\;{\rm{ = }}\;14.6\;{\rm{kJ}}\]