Ask question

Ask question

All categories

11 days ago

8

18. How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C

so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat.

1 answer:

Anarel [2.9K]11 days ago

3 0

Answer:

The amount of water needed to mix with the solution is 170.27 mL

Explanation:

Provided details:

Water temperature ( $T_{1}$ ) = $23^{o}C$

density of water (ρ) = 1.00 g/mL

Coffee temperature ( $T_{1}$ ) = $95^{o}C$

Coffee volume ( $V_{2}$ ) = 180 mL

Mixed Temperature ( $T_{mix}$ ) = $60^{o}C$

To calculate the heat, we apply the formula:

Q = m x c x ΔT

where

m = mass (g)

c = specific heat (J/ $g^{o}C$ )

ΔT = temperature change ( $^{o}C$ )

In this mixed solution, the water's heat ( $Q_{1}$ ) must equal the coffee's heat ( $Q_{2}$ ). Thus,

$Q_{1}$ = $Q_{2}$

$m_{1}$ x $c_{1}$ x Δ $T_{1}$ = $m_{2}$ x $c_{2}$ x Δ $T_{2}$

where

$m_{1}$ is the water’s mass

is the coffee’s mass $m_{2}$

refers to the specific heat of water

$c_{1}$ and

is the specific heat of coffee $c_{2}$

Assuming the specific heats of both coffee and water are identical, therefore,

$c_{1}$ = $c_{2}$ , can be canceled in the equation.

Thus,.

$m_{1}$ x Δ $T_{1}$ = $m_{2}$ x Δ $T_{2}$

We know that

ρ = $\frac{m}{V}$

m = ρ x V, substitute this in the equation:

ρ $_{1}$ x V $_{1}$ x Δ $T_{1}$ = ρ $_{2}$ x V $_{2}$ x Δ $T_{2}$

$V_{1}$ is the volume of water

Since coffee and water share the same density, hence, we can eliminate the formula

V $_{1}$ x Δ $T_{1}$ = V $_{2}$ x Δ $T_{2}$

Thus, V $_{1}$ = (V $_{2}$ x Δ $T_{2}$ ) / Δ $T_{1}$ V $_{1}$ = V $_{2}$ x ( $\frac{(T_{2} - T_{mix}) }{(T_{mix} - T_{1})}$

V $_{1}$ = 180 mL x $\frac{(95-60)^{o}C}{(60-23)^{o}C}$

V $_{1}$ = 180 mL x $\frac{(35)^{o}C}{(37)^{o}C}$

V $_{1}$ = 170.27 mL

You might be interested in

Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

eduard [2782]

Clarification:

The pertinent information is outlined as follows.

m = 10.0 kg = 10,000 g (since 1 kg = 1000 g)

Starting temperature of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Starting temperature of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

Therefore, the heat lost by block 1 equals the heat received by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

It's important to convert the temperature into Kelvin as (50 + 273) K = 323 K.

Additionally, the relationship between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Next, determine the entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, the entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Thus, the total entropy is the sum of the entropy changes of both blocks.

= -554.12 J/K + 647.49 J/K $\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= 93.37 J/K

In conclusion, for this reaction, the outcome is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

6 0

28 days ago

Why does 5060 have three significant figures?

Anarel [2989]

5060 has three significant figures: Below is the clarification

Explanation:

Significant figures

Significant figures (also referred to as significant digits and decimal places) in a number are those digits that carry substantial meaning.

These include all digits except: leading zeros.

Guidelines for determining significant figures

1. All non-zero digits are counted as significant. For instance, the number 23 has two significant figures.

2. Zeros located between two non-zero digits are significant; for example, 202.1201 contains seven significant figures.

3. Zeros preceding the significant figures are not significant. For example,.000021 has two significant figures, with zeros being non-contributory.

4. Zeros following the significant figures are significant.

This explains why the number 5060 has three significant figures.

3 0

1 month ago

32.7 grams of water vapor takes up how many liters at standard temperature and pressure (273 K and 100 kPa)?

alisha [2963]

At standard temperature and pressure, it is established that 1 mole of gas has a volume of 22.4 liters.

According to the periodic table:
the molar mass of oxygen is 16 g
and the molar mass of hydrogen is 1 g
Hence, the molar mass of water vapor is calculated as 2(1) + 16 = 18 g

Thus, 18 g of water occupies 22.4 liters, therefore:
the volume for 32.7 g is (32.7 x 22.4) / 18 = 40.6933 liters

5 0

1 month ago

Compute 4.659×104−2.14×104. Round the answer appropriately.

lions [2927]

Answer: 25,200.

Explanation:

1) Starting with: 4.659 × 10⁴ - 2.14 × 10⁴

2) Significant figures need to be considered.

Because the powers are equal (10⁴), the decimal values can be subtracted directly. However, it is essential to first check significant figures and the count of decimal places.

3) The figure 4.659 × 10⁴ has four significant digits (4, 6, 5, and 9), whereas 2.14 × 10⁴ contains three significant digits (2, 1, and 4).

4) When adding or subtracting values with a different number of decimal places, the answer must reflect the same number of decimal places as the number with the least amount of decimal precision.

5) Prior to performing subtraction, it is necessary to round the numbers to the least decimal places. Given that 2.14 has two decimal places and 4.659 has three, round 4.659 to 4.66.

6) Now perform the subtraction 4.66 - 2.14 = 2.52

7) Then multiply by the power of 10: 2.52 × 10⁴ = 25,200. This is the final answer.

5 0

26 days ago

Read 2 more answers

Which equation represents a spontaneous reaction?

Anarel [2989]

Answer:

D) Mn + Ni2+ ⇒ Mn2+ + Ni

Explanation:

A spontaneous process can occur in a specific direction without requiring any energy input from external sources. Such reactions happen naturally. In these spontaneous processes, the entropy change is positive (ΔS), the enthalpy change is negative (ΔH), and most importantly, ΔG (the change in free energy) is negative.

To identify which reaction is spontaneous, we analyze the electrode potentials of the involved species. The species with a more negative reduction potential can displace the other from its aqueous solution. In this case, since the reduction potential for Mn^2+ is -1.19 V compared to nickel's -0.25 V, manganese will thus naturally displace Ni^2+ from solution as indicated in the solution above.

5 0

27 days ago