An 8.6 grams piece of aluminum, heated to 100°c, is placed in a coffee cup (styrofoam) calorimeter that contains 402.4 grams of water at 25°c. the specific heat of water is 4.18 j/g°c and the specific heat of aluminum is 0.900 j/g°c.
what will be the final equilibrium temperature for both the aluminum and the water? (assume no heat is lost to the surroundings and the coffee cup absorbs a negligible amount of heat).

The final equilibrium T° will be 24.65°C

Explanation:

A typical excersise of calorimetry, to apply this formula:

Q =  m . C . ΔT

As no heat is lost to the surroundings, the heat that loses the aluminum and gains the water will be the same.

mass Al . C Al . ΔT = mass water . CH2O . ΔT

ΔT = T°final - T°initial

8.6 g . 0.900 J/g°C . (T°final - 100°C) = 402.4 g . 4.18J/g°c .  (T°final - 25°C)

7.74 J/°C T°final - 774J = 1682.03 T°final - 42050.8J

-774 J + 42050.8J =  1682.03 T°final - 7.74 J/°C T°final

41276.8 J = 1674.29 J/°C T°final

41276.8 J /1674.29 °C/J =T°final

24.65°C = T°final

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