2grams of a substance was dissolved in 75.0 g of water resulting in a temperature decrease from 23.1 ⁰c to 18.7 ⁰c. the specific heat of the solution is 4.18 j/g·⁰c. calculate the amount of heat associated with the water.

1379.4 Joules

Therefore;

Quantity of heat = Mass × specific heat capacity × Change in temperature

Q = mcΔT

In this case;

The substance dissolved in water gained heat while water lost heat energy.

Thus, Heat gained by the substance = heat lost by water

Heat associated with the water  

Mass of water = 75 g

Change in temperature = 4.4°C

Specific heat capacity =  4.18 J/g·⁰C

Heat = mcΔT

        = 75 g × 4.18 J/g·⁰C × 4.4 °C

        =1379.4 Joules

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

58750.08068 is the answer i think