A 1.35-gram sample of compound X (of MW

88.0 g/mol) was burned in a bomb calorimeter
containing 1700 g of water. A temperature
rise of 0.96◦C was observed. ∆Erxn for this reaction
is −475 kJ/mol of X. What is the heat
capacity of the calorimeter hardware (not the
water)?
Answer in units of J/
â—¦C.

massX*/molmassX *Hreactionx=masswaaterkg*specificHeatwater*temmp rise

make certain you use for water specific heat J/ kg units. watch units.

To find the heat capacity of the calorimeter hardware, we need to use the equation:

q = mc∆T

where:
q is the heat transferred
m is the mass of the substance (in this case, the calorimeter hardware)
c is the specific heat capacity of the substance
∆T is the change in temperature

In this case, we know the values for q, m, and ∆T, and we need to find the value for c.

The heat transferred, q, can be calculated using the equation:

q = ∆E + q_water

where:
∆E is the change in internal energy of the system
q_water is the heat absorbed or released by the water

Given that ∆E for the reaction is -475 kJ/mol (negative because it is a release of energy) and that one mole of compound X has a molecular weight of 88.0 g/mol, we can calculate ∆E for the sample of compound X burned in the calorimeter:

∆E = -475 kJ/mol × (1.35 g / 88.0 g/mol)

∆E = -7.27 kJ

Now, we can calculate q_water using the equation:

q_water = m_water × c_water × ∆T

where:
m_water is the mass of water
c_water is the specific heat capacity of water

Given that the mass of water, m_water, is 1700 g and the change in temperature, ∆T, is 0.96 °C, and the specific heat capacity of water, c_water, is 4.18 J/(g·°C), we can calculate q_water:

q_water = (1700 g) × (4.18 J/(g·°C)) × (0.96 °C)

q_water = 6532.16 J

Now, we can calculate q:

q = ∆E + q_water

q = -7.27 kJ + 6532.16 J

q = -7.27 kJ + 6.53 kJ

q = -0.74 kJ

Finally, we can rearrange the equation q = mc∆T to solve for c:

c = q / (∆T × m)

c = -0.74 kJ / (0.96 °C × m)

Given that we want the answer in units of J/°C, we need to convert kJ to J:

c = -740 J / (0.96 °C × m)

Therefore, the heat capacity of the calorimeter hardware (not the water) is -740 J/°C, where m is the mass of the calorimeter hardware.