The initial temperature of the water in a constant-pressure calorimeter is 24°c. a reaction takes place in the calorimeter, and the temperature rises to 87°c. the calorimeter contains 367 g of water, which has a specific heat of 4.18 j/(g·°c). calculate the enthalpy change (δh) during this reaction.

4739.86 J/mol.

Explanation:

The amount of heat released to water = Q = m.c.ΔT.

where, m is the mass of water (m = 367.0 g).

c is the specific heat capacity of water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 87.0°C - 24.0°C = 63.0°C).

∴ The amount of heat absorbed by released to water = Q = m.c.ΔT = (367.0 g)(4.18 J/g°C)(63.0°C) = 96645.78 J.

ΔH = Q/n.

n = mass/molar mass = 367 g/18 g/mol = 20.39 mol.

∴ ΔH = Q/n = - (96645.78 J)/(20.39 mol) = 4739.86 J/mol.

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the answer is c