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liubo4ka [24]
3 years ago
8

how much heat, in terms in q, would it take to produce the change in temperature indicated in the picture? what is your reasonin

g?​

Chemistry
1 answer:
STALIN [3.7K]3 years ago
4 0

Answer:

1. q.

2. 2q.

3. 3q.

4. 6q.

Explanation:

We'll begin by calculating the specific heat capacity of the liquid. This can be obtained as follow:

Mass (m) = 25 g

Change in temperature (ΔT) = 20 °C

Heat (Q) = q

Specific heat capacity (C) =?

Q = MCΔT

q = 25 × C × 20

q = 500C

Divide both side by 500

C = q/500

C = 2×10¯³ qg°C

Therefore, the specific heat capacity of liquid is 2×10¯³ qg°C

Now, we shall determine the heat required to produce the various change in temperature as follow:

2. Mass (m) = 50 g

Change in temperature (ΔT) = 20 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 50 × 2×10¯³ × 20

Q = 2q.

Therefore, the heat required is 2q.

3. Mass (m) = 25 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 25 × 2×10¯³ × 60

Q = 3q.

Therefore, the heat required is 3q.

4. Mass (m) = 50 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) = 2×10¯³ qg°C

Heat (Q) =?

Q = MCΔT

Q = 50 × 2×10¯³ × 60

Q = 6q.

Therefore, the heat required is 6q.

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#1

  • Volume=15mL
  • Mass=40.5g

\\ \sf\longmapsto Density=\dfrac{Mass}{Volume}

\\ \sf\longmapsto Density=\dfrac{40.5}{15}

\\ \sf\longmapsto Density=2.7g/mL

#2

  • Mass=306g
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\\ \sf\longmapsto Density=\dfrac{306}{22.5}

\\ \sf\longmapsto Density=13.6g/mL

#3

  • Density=0.789g/mL
  • Volume=200mL

\\ \sf\longmapsto Mass=Density(Volume)

\\ \sf\longmapsto Mass=0.789(200)

\\ \sf\longmapsto Mass=157.8g

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\\ \sf\longmapsto V=LBH

\\ \sf\longmapsto V=8.4(5.5)(4.6)

\\ \sf\longmapsto V=212.5cm^3

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\\ \sf\longmapsto Density=\dfrac{1896}{212.5}

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Read 2 more answers
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