   Chapter 6, Problem 68E

Chapter
Section
Textbook Problem

# In a coffee-cup calorimeter, 1.60 g NH4NO3 is mixed with 75.0 g water at an initial temperature of 25.00°C. After dissolution of the salt, the final temperature of the calorimeter contents is 23.34°C. Assuming the solution has a heat capacity of 4.18 J/°C · g and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of NH4NO3 in units of kJ/mol.

Interpretation Introduction

Interpretation: The enthalpy change for dissolving the NH4NO3 salt in Water has to be calculated.

Concept Introduction: The heat capacity C is defined as the ratio of heat absorbed and temperature change. It can be given by,

C = Heatabsorbed(inJ)Temperature(in°C)

• The required heat of a one gram of substance raise to its temperature by one degree celsius is called specific heat capacity.

Absorbed heat (J) = Specific heat capacity ×Temperature change(°C)×mass ofsubstance (g) (1)

For in above equation heat is:

q= S×M×T (2)

S is specific heat capacity (J°C.g)

M is mass of sample ( g )

T is temperature change ( °C )

For the process no heat loss to the surroundings means then the heat is

(absorbed) =-q (released) (3)

Explanation

Explanation

Given data:

Water weight = 75g

Mass of NH4NO3 = 1.60g

Initial temperature of water = 25°C

Final temperature of solution =23.34°C

Specific heat capacity of the solution = 4.18 JCg-1

The final weight of solution is 75g +1.60g = 76.60g

To determine: Heat lost by solution

Heat lost by solution is equal to heat gained by NH4NO3

Heat lost by solution = 4.18×76.60×(25-23.34)

= 532 J

ΔH is in J/g units is = 532J1

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