The molar heat of combustion of compound C is 1,250 kJ mol-1. If I were to burn 0.115moles of this compound in a bomb calorimeter with a reservoir that holds 2.50 dm3ofwater, what would the expected temperature increase be?
Question
The molar heat of combustion of compound C is 1,250 kJ mol-1. If I were to burn 0.115moles of this compound in a bomb calorimeter with a reservoir that holds 2.50 dm3ofwater, what would the expected temperature increase be?
Solution
To solve this problem, we need to use the formula q = mcΔT, where q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
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First, we need to calculate the total heat produced by the combustion. The molar heat of combustion is the heat produced by one mole of a substance. So, for 0.115 moles, the heat produced (q) would be 0.115 moles * 1250 kJ/mol = 143.75 kJ.
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Next, we need to convert the volume of water to mass. Since the density of water is approximately 1 g/cm3, and 1 dm3 is equal to 1000 cm3 (or 1000 g), the mass of the water (m) is 2.50 dm3 * 1000 g/dm3 = 2500 g.
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The specific heat capacity (c) of water is approximately 4.18 J/g°C. However, our heat is in kJ, so we need to convert it to J by multiplying by 1000: 4.18 J/g°C = 4.18 kJ/kg°C.
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Now we can rearrange the formula to solve for ΔT: ΔT = q / (mc) = 143.75 kJ / (2500 g * 4.18 kJ/kg°C) = 0.0137°C.
So, the expected temperature increase would be approximately 0.0137°C.
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