Asample of an unknown substance has a mass of .158 kg if 2,510 j of heat is required to heat the substance from 32°c to 61°c what is the specific heatt of the substance
.171
.548
15.9
86.6

0.548J/g°C

Explanation:

Q = s × m × DeltaT

Q = Heat (J)

S = Specific Heat Capacity

M = mass (g)

DeltaT = Change in temperature (°C)

0.158Kg x 1000 = 158g

2.510J = s x 158g x (61°C-32°C)

2.510J/(158g x 29°C) = s

S = 0.54779 J/g°C

S = 0.548 J/g°C

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