Question

Calculate the change in internal energy of 2kg of water at 90 degree celcius when it is changed to 330m3 of steam at 100oC. The whole process occurs at atmospheric pressure. The latent heat of vaporization of water is 226106 J/kg.
4.27 MJ
3.43 kJ
45.72 mJ
543.63 J

18 Tensile strain is mathematically expressed as:
Force/Area
initial length/extension
extension/initial lenght
Stress + initial lenght

19 A certain resistance thermometer at triple point of water has resistance of
152.0Ω
. What is the temperature of the system in degrees celcius when the resistance of the thermometer is
230.51Ω
?
414.2o
C
141.0o
C
253.2o
C
80.4o
C

Answer 1

Answer on Question #52082, Physics, Field Theory

Calculate the change in internal energy of $2\mathrm{kg}$ of water at 90 degree celcius when it is changed to $330\mathrm{m}^3$ of steam at $100\mathrm{oC}$. The whole process occurs at atmospheric pressure. The latent heat of vaporization of water is $226106\mathrm{J/kg}$.

$4.27\mathrm{MJ}$

$3.43\mathrm{kJ}$

$45.72\mathrm{mJ}$

$543.63\mathrm{J}$

Solution

$$Q_w = m_{\text{water}} c \cdot \Delta T = 2\mathrm{kg} \cdot 4187\mathrm{J}/\mathrm{kg} \cdot K \cdot 10K = 83740\mathrm{J}$$

where $m_{\text{water}} = 10\mathrm{kg}$ is the mass of the water; $\Delta T = 10K$ is the change of temperature; $c = 4187\mathrm{J}/\mathrm{kg} \cdot K$ is the specific heat capacity

$$Q_g = \rho_g \cdot V \cdot L = 330\mathrm{m}^3 \cdot 0.6\mathrm{kg}/\mathrm{m}^3 \cdot 226106\mathrm{J}/\mathrm{kg} = 4.47 \cdot 10^6\mathrm{J}$$

Then $Q = Q_W + Q_g = 4.48 \cdot 10^7\mathrm{J}$

**Answer**: $Q = 4.48 \cdot 10^7\mathrm{J}$

18 Tensile strain is mathematically expressed as:

- Force/Area

- initial length/extension

- extension/initial lenght

- Stress + initial length

**Answer**: extension/initial lenght

19 A certain resistance thermometer at triple point of water has resistance of $152.0\Omega$. What is the temperature of the system in degrees celcius when the resistance of the thermometer is $230.51\Omega$?

$414.2{}^{\circ}\mathrm{C}$

$141.0{}^{\circ}\mathrm{C}$

$253.2{}^{\circ}\mathrm{C}$

$80.4{}^{\circ}\mathrm{C}$

Solution

Fig.1

The temperature is

$$T (R) = (273.15K) \cdot \frac{R}{R_{3}} = (273.15K) \cdot \frac{230.51}{152.0} = 414.2K$$

Answer: $T(R) = (273.15K) \cdot \frac{R}{R_{3}} = 414.2K$

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