1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
slamgirl [31]
3 years ago
6

A student constructs a coffee cup calorimeter and places 50.0 mL of water into it. After a brief period of stabilization, the te

mperature of the water in the calorimeter is determined to be 22.6 °C. To this is added 50.0 mL of water that was originally at a temperature of 54.5 °C. A careful plot of the temperatures recorded after this established the temperature at T0 was 30.31 °C. What is the calorimeter constant in J/°C for this calorimeter?
Chemistry
1 answer:
Grace [21]3 years ago
5 0

Answer:

The calorimeter constant is  = 447 J/°C

Explanation:

The heat absorbed or released (Q) by water can be calculated with the following expression:

Q = c × m × ΔT

where,

c is the specific heat

m is the mass

ΔT is the change in temperature

The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.

The heat absorbed by the calorimeter (Q) can be calculated with the following expression:

Q = C × ΔT

where,

C is the calorimeter constant

The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).

Qabs + Qrel = 0

Qabs = - Qrel

Qcal + Qw₁ = - Qw₂

Qcal = - (Qw₂ + Qw₁)

Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)

Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) +  (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]

Ccal  = 447 J/°C

You might be interested in
A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther
sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )

1. Calculate the thermometer readings after  0.5 min and 1 min

(a) After 0.5 min

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}

(b) After 1 min

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

  t = t - 1, because the cooling starts 1 min late

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0
3 years ago
Land forms on earth change everyday. Some are worn away and even destroyed. Some are built up and even formed anew. Processes th
kherson [118]
Missing part of the question. Please add which are the process to be classified.
5 0
3 years ago
Read 2 more answers
Which statement about Niels Bohr’s atomic model is true?
jenyasd209 [6]

Answer:

Each orbit has a specific energy level.

Would you mind marking it the brainliest:).

4 0
3 years ago
If 0.60 L of a solution contains 6.6 g of NaBr, what is its molar concentration?
-BARSIC- [3]

Answer:

Molarity = moles ÷ liters

to get moles of NaBr divide grams of NaBr by its molar mass (mass of Na + mass of Bromine)

Na = 22.989769

Br = 79.904

molar mass of NaBr = 102.893769

6.6g ÷ 102.893769 = 0.064143826 moles of NaBr

0.064143826 moles ÷ 0.60 liters = 0.1069 molar concentration or 11 %

7 0
3 years ago
Equal masses of substance A and substance B are at 25°C. Substance A has a lower specific heat capacity than substance B. If 50
liraira [26]
Dang bro that stuff is really hard I’m defiantly on a lower grade lol
6 0
3 years ago
Other questions:
  • Match the term with its description. match term definition acid
    6·1 answer
  • When the equation is correctly balanced, the coefficient of aluminum chloride will be ___.
    15·1 answer
  • the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal
    5·2 answers
  • Which liquid would be the most difficult to raise or lower the temperature of
    5·1 answer
  • Which of the following atoms is likely to transfer (and lose) an electron when it forms an ion? Please choose the correct answer
    10·1 answer
  • We could avoid a large increase in temperature if greenhouse emissions peaked by the year of ?
    12·2 answers
  • Complete el crucigrama sobre conceptos del agua.
    11·1 answer
  • Which of this is the method by which fire extinguisher works?
    7·2 answers
  • How many kilograms of water must be added to 6.07 grams of oxalic acid (H2C2O4) to make a 0.025 m solution?
    14·1 answer
  • For each ionic compound formula, identify the main group to which X belongs: (a) XF2;
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!