Basically, this is a unit conversion problem. Starting with grams of sulfur, convert to moles of sulfur, and finally to kJ of heat released. Note that the given enthalpy of reaction, , is the amount of heat released when two moles of sulfur react with three moles of oxygen.

Note that a negative enthalpy of reaction means the process is exothermic and releases heat, while a positive enthalpy of reaction would mean the process is endothermic and absorbs heat.

Since the system is isolated, the amount of heat transferred away from the unknown substance must equal the heat transferred to the water. To calculate these heats, use 

For the water, .

For the unknown substance, 

Set these values equal to get c_x = 2.39 J/g^oC, approximately 2.4 J/g^oC.

There are three distinct steps in this transformation: 1) heating the ice from -30^oC to 0^oC, 2) melting the ice at 0^oC, and 3) heating the water from 0^oC to 40^oC.

When there is a temperature change, we use , and when there is melting or freezing, we use .

1) 

2) 

3) 

Adding these three pieces together gives the total enthalpy change, which is equal to change in energy.

In a closed system, energy and work can be exchanged with the surrounding area but matter is contained. Consider, for example, a beaker of ice. Heat from the surroundings it transferred to the ice, causing it to melt, but there is no exchange of matter. No external compounds react with the ice and none of the ice leaves the beaker.

In an open system, both heat and matter can be exchanged. In an isolated system neither heat, nor matter is exchanged.

This process can be divided into three stages: the raising of the temperature of the ice, the melting of the ice into water, and the raising of the temperature of the water. The first and third step use the specific heat of the substance to calculate the heat required, while the second step uses the heat of fusion.

Step 1 and 3:

Step 2:

We know the mass of the water, the change in temperature, and the constant values. Using these values, we can sum the heat required for each step to get our final answer.

There are two processes requiring added heat in this problem:

1. Raising the temperature of the liquid water from  to  (use )

2. Boiling the water at a constant temperature of  (use )

To use either of these equation, we need to find the mass of the water using the relation between mass, density, and volume.

Use this mass with the given specific heat and temperatures to find the heat for part 1 of the process.

Then, use the mass with the given heat of vaporization to find the energy needed to convert the water to water vapor.

Sum the energies for step 1 and step 2.

This is the total amount of energy needed from the burning wood. Use stoichiometry to find the grams of wood needed to produce this amount of energy.

If heat is transferred from the metal object to the water, that means that the temperature of the metal must be greater than that of the water. We must find the answer choice where this is the case. Since the temperatures are given in different units, they must be converted to the same unit for comparison purposes.

Only one answer option results in the metal having a greater temperature than the water:

In this problem, we need to track the transfer of heat from the aluminum to the water. Since the heat acquired by the water is equal to the heat given off by the aluminum, we can set their equations equal to each other; however, in order to avoid a negative number, aluminum's change in temperature will be set as the initial temperature minus the final temperature. This results in the equation below.

When a specific heat capacity is given, we typically use the equation . Since we know all of the factors except for the change in temperature, we can simply solve for .

The equation for enthalpy of reaction is:

Given our chemical reaction and the enthalpies of formation, we can find the enthalpy of reaction.

First, find the total enthalpy for the products.

Then, find the total enthalpy for the reactants.

Since the oxygen is elemental, its heat of formation is zero.

Return to the original equation to calculate the final enthalpy of reaction.