aluminum has a specific heat of 0.897 j/g°C. how much heat is needed to raise the temperature of a 79 gram piece of aluminum 28°C?​

Approximately 2000 J.

General Formulas and Concepts:

Thermodynamics

Specific Heat Formula: q = mcΔT

Explanation:

Step 1: Define

Identify variables

[Given] c = 0.897 J/g °C

[Given] m = 79 g

[Given] ΔT = 28°C

[Solve] q

Step 2: Solve for q

Step 3: Check

Follow sig fig rules and round. We are given 2 sig figs as our lowest.

1984.16 J ≈ 2000 J

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

i'm pretty sure it is letter c.) car