How much energy (in joules) is required to convert 129 grams of ice at −23.0 °c to liquid water at 18.0 °c?

(the specific heat of ice = 2.10 j/g°c; the specific heat of water = 4.184 j/g°c; the heat of fusion = 333 j/g°c).

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = 58902.9 joules

Explanation:

This is a calorimetry problem:

Q = m . C . ΔT

Q = heat; m = mas; C is the specific heat and

ΔT = Final T° - Initial T°

Q = C lat . m

Q = Heat

m = mass

C lar = Latent heat of fusion

First of all we calculate the heat for ice, before it takes the melting point. (from -23°C  to 0°C)

Q = 129 g . 2.10 J/g°C . (0°C - (-23°C)

Q = 129 g . 2.10 J/g°C . 23°C → 6230.7 joules

Then, the ice has melted. To be melted and change the state it required:

Q = C lat . m

Q = 333 J/°C . 129 g → 42957 joules

And in the end, we have water that changed its T° from O°C to 18°C

Q = 129 g . 4.184 J/g °C . (18°C - 0°C)

Q = 9715.2 Joules

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = 58902.9 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