? Question

An isolated insulated cylinder contains 15.40 grams of neon gas and a 62.30-gram bronze medal in thermal contact. When

the bronze medal was put into the cylinder, its temperature was 72.30°C. The final temperature of both the medal and the

neon gas is 66.30°C. Neon's specific heat capacity is 1.047 joules/gram degree Celsius, and bronze's specific heat capacity is

0.3770 joules/gram degree Celsius. What was the initial temperature of the neon?

1.1°C

Explanation:

Heat lost by bronze is

H = mc(T2-T1)

H is the heat,

m is mass of bronze 62.3 gram

T1 = initial temperature = 72.30°C

T2 = final temperature = 66.30°C

c is heat capacity, 1.047J/g-°C

Therefore,

H = 62.3 x 1.047 x (66.3 - 72.3)

H = -391.37 Joules

The -ve sign indicate that the heat is lost to the neon.

This amount of heat is gained by the neon.

For the neon,

m = 15.40 gram

c = 0.3770 J/g-°C

T1 =?

T2 = 66.3°C

Equating to the heat gained we have

391.37 = 15.40 x 0.377 x (66.3 - T1)

391.37 = 5.8058 x (66.3 - T1)

67.4 = 66.3 - T2

T1 = 67.4 - 66.3 = 1.1°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

it takes on an electronical energy