Title: Défaut pour Quiz Calorimetry - 2

La catégorie par défaut pour les questions partagées dans le contexte « Quiz Calorimetry - 2 ».

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Q: Let's consider the following temperatures: T_1=70°C, T_2=52°C.

Calculate :

- in K    T_1 = [...] K    T_2 = [...] K

- in °C  T_1-T_2 = [...] °C

- in K    T_1-T_2 = [...] K

- in °C   \( {T}_{1}^{2} \)-\[ {T}_{2}^{2} \] = [...] °C

- in K    \( {T}_{1}^{2} \)-\[ {T}_{2}^{2} \] = [...] K
A: 343; 325; 18; 18; 2196; 12024
Feedback: - in K T1 = 343 K T2 = 325 K - in °C T1-T2 = 18°C - in K T1-T2 = 18 K - in °C T_1^2-T_2^2 = 2196°C - in K T_1^2-T_2^2 = 12024 K

Q: Let's consider the following temperatures: T_1=70°C, T_2=52°C.

Calculate :

- in K    T_1 = [...] K    T_2 = [...] K

- in °C  T_1-T_2 = [...] °C

- in K    T_1-T_2 = [...] K

- in °C   \( {T}_{1}^{2} \)-\[ {T}_{2}^{2} \] = [...] °C

- in K    \( {T}_{1}^{2} \)-\[ {T}_{2}^{2} \] = [...] K
A: 343; 325; 18; 18; 2196; 12024
Feedback: Reminder: T(K)=T(°C)+273 Solution: - in K T1 = 343 K T2 = 325 K - in °C T1-T2 = 18°C - in K T1-T2 = 18 K - in °C T_1^2-T_2^2 = 2196°C - in K T_1^2-T_2^2 = 12024 K

Q: Drop the phases changes on their appropriate location :
- Zone 1 | Deposition
- Zone 2 | Melting
- Zone 3 | Vaporization
- Zone 4 | Freezing
- Zone 5 | Sublimation
- Zone 6 | Condensation

Q: If we heat two different liquids on a heat plate for a same duration, the two liquids will have the same final temperature :
- True
- False
A: False
Feedback: If the liquids are different, they have different specific heat capacities c. Therefore, according to the equation of the sensible heat Q=m.c.(Tf-Ti), if Q is the same, then the final temperature Tf will be different.

Q: If I mix 1 L of water at 20°C and 2 L of water at 20°C, the final temperature will be :
- 20°C
- 40°C
A: 20°C
Feedback: The temperatures of two bodies do not add.

Q: In an experiment, a sample of copper has its temperature that dropped from 70°C to 40°C.

Did the sample receive a quantity of heat or give a quantity of heat ?
- Receive
- Give
A: Give
Feedback: If the temperature drops, then it means that the copper gives a quantity of heat. By convention, the quantity of heat is negative.

Q: The latent heat of vaporization of a refrigerant is 232 kJ/kg.

What is its latent heat of condensation expressed in kJ/kg ?
A: -232
Feedback: The latent heat of condensation is the opposite of the latent heat of vaporization, therefore it is -232 kJ/kg.

Q: What is the quantity of heat necessary to melt 10 kg of ice ?
A: 3340000
Feedback: The quantity of heat necessary to melt 10 kg of ice is Q = m'.L=10 kg x 334 kJ/kg = 3 340 kJ= 3 340 000 J.