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vesna_86 [32]
3 years ago
7

A 25 gram(m) metal ball is heated to 200C(delta T) with 2330 Joules(q) of energy. What is the specific heat of the metal?

Chemistry
2 answers:
Dominik [7]3 years ago
6 0

Answer:

The specific heat of the metal is 0.466 \frac{J}{g*C}

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

The equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

  • Q= 2330 J
  • c= ?
  • m= 25 g
  • ΔT= 200 °C

Replacing:

2330 J= c*25 g* 200 °C

Solving:

c=\frac{2330 J}{25 g* 200 C}

c=0.466 \frac{J}{g*C}

<u><em>The specific heat of the metal is 0.466 </em></u>\frac{J}{g*C}<u><em></em></u>

marin [14]3 years ago
4 0

Answer:

Cp=0.466\frac{J}{g^oC}

Explanation:

Hello,

In this case, we compute the heat as shown below:

Q=mCp\Delta T

Thus, since we know the heat, mass and change in temperature, by solving for the heat capacity we obtain:

Cp=\frac{Q}{m\Delta T} =\frac{2330J}{25g*200^oC} \\\\Cp=0.466\frac{J}{g^oC}

Best regards.

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Explanation:

Ideal Gases Law can useful to solve this

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Consider the data presented below. time (s) 0 40 80 120 160 moles of a 0.100 0.067 0.045 0.030 0.020 part a part complete determ
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To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows 
time         0         40        80       120       160
moles    0.100   0.067  0.045    0.030    0.020


Q1)
for the first 40 s change of moles ;
      = -d[A] / t
      = - (0.067-0.100)/40s
      = 8.25 x 10⁻⁴ mol/s
for the next 40 s
      = -(0.045-0.067)/40
      = 5.5 x 10⁻⁴ mol/s
the 40 s after that
      = -(0.030-0.045)/40 s 
     = 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction

rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ   -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction

Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation. 
Since this is a first order reaction,
b = 1
therefore the reaction is 
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
   = 8.25 x 10⁻³ s⁻¹

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