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antiseptic1488 [7]
3 years ago
5

#1: Which of the following is the abbreviation for a unit of energy? A. K / B. °C/ C. W / D. cal............... #2: A 200 g bloc

k of a substance requires 1.84 kJ of heat to raise its temperature from 25°C to 45°C. Use the table attached to identify the substance. A. iron/ B. aluminum/ C. gold/ D. copper.....................#3: In a calorimeter, the temperature of 100 g of water decreased by 10°C when 10 g of ice melted. How much heat was absorbed by the ice? A. 418 kJ / B. 100 kJ / C. 10 J / D. 4.18 kJ ..................#4: The amount of heat needed to raise the temperature of 50 g of a substance by 15°C is 1.83 kJ. What is the specific heat of the substance? A. 2.44 J/g-°C / B. 2.22 J/g-°C / C. 2.13 J/g-°C / D. 2.05 J/g-°C ................#5: In a calorimeter, 3.34 kJ of heat was absorbed when 10 g of ice melted. What is the enthalpy of fusion of the ice? A. 6.68 J/g / B. 334 J/g / C. 6.68 kJ/g/ D. 334 kJ/g
Chemistry
2 answers:
stealth61 [152]3 years ago
8 0
Among the choices, the unit of energy is calories. Answer in 1) is D. In 2) we are given with te mass , heat and temperature change. we just need to get the heat capacity and compare it with the following metals. The calculated heat capacity is 0.46 kJ/kg K. The answer is A. iron. In 3) we can compute the heat absorbed by the formula ΔH=mCpΔT. Cp of water is 4.18 J/g K. Answer of 3) is D. In 4) the formula used in Cp=ΔH/mΔT. Answer in 4) is A. The heat of enthalpy of fusion of ice is 80 cal/g. We convert this to J/g. Answer  in 5) is B.334 J/g.
Ivenika [448]3 years ago
7 0

Answer:

1. D. cal......

2.A. iron

3. D

4.2.44j/g°C   A

5,Lf=334J/g   B

Explanation:

1: Which of the following is the abbreviation for a unit of energy? A. K / B. °C/ C. W / D. cal...............

calorie is the unit of energy

#2: A 200 g block of a substance requires 1.84 kJ of heat to raise its temperature from 25°C to 45°C. Use the table attached to identify the substance. A. iron/ B. aluminum/ C. gold/ D. copper.....................

Q=mcdt

1840=0.2*C*(45-25)

C=460J/KgK

if the specific heat capacity is the above then he substance is iron

#3: In a calorimeter, the temperature of 100 g of water decreased by 10°C when 10 g of ice melted. How much heat was absorbed by the ice? A. 418 kJ / B. 100 kJ / C. 10 J / D. 4.18 kJ .................

Q=mcdT

Q=0.1*10*4180

Q=4180j. answer D

.#4: The amount of heat needed to raise the temperature of 50 g of a substance by 15°C is 1.83 kJ. What is the specific heat of the substance? A. 2.44 J/g-°C / B. 2.22 J/g-°C / C. 2.13 J/g-°C / D. 2.05 J/g-°C ................

Q=mcdT

1830=50/1000*C*15

C=2440j/kg/k

change it to j/g°C

2.44j/g°C   A

#5: In a calorimeter, 3.34 kJ of heat was absorbed when 10 g of ice melted. What is the enthalpy of fusion of the ice? A. 6.68 J/g / B. 334 J/g / C. 6.68 kJ/g/ D. 334 kJ/g

Q=mLf

Lf=enthalpy of fusion

3340/10=Lf

Lf=334J/g   B

Enthalpy of fusion quantity of heat to convert 1 unit mass of a solid to liquid without any noticeable change in temperature.

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1. The heat of fusion for the ice-water phase transition is 335 kJ/kg at 0°C and 1 bar. The density of water is 1000 kg/m3 at th
vodomira [7]

Answer:

Expression for the change of melting temperature with pressure..> T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa), Freezing Point = 0°C

Explanation:

Derivation from state postulate

Using the state postulate, take the specific entropy,  , for a homogeneous substance to be a function of specific volume  and temperature  .

ds = (partial s/partial v)(t) dv + (partial s/partial T)(v) dT

During a phase change, the temperature is constant, so

ds = (partial s/partial v)(T)  dv

Using the appropriate Maxwell relation gives

ds = (partial P/partial T)(v) dv

s(β) – s(aplαha) = dP/dT (v(β) – v(α))

dP/dT = s(β) – s(α)/v(β) – v(α) = Δs/Δv

Here Δs and Δv are respectively the change in specific entropy and specific volume from the initial phase α to the final phase β.

For a closed system undergoing an internally reversible process, the first law is

du = δq – δw = Tds - Pdv

Using the definition of specific enthalpy, h and the fact that the temperature and pressure are constant, we have

du + Pdv = dh Tds,

ds = dh/T,

Δs = Δh/T = L/T

After substitution of this result into the derivative of the pressure, one finds

dp/dT = L/TΔv

<u>This last equation is the Clapeyron equation.</u>

a)

(dP/dT) = dH/TdV => dP/dlnT = dH/dV

=> dP/dlnT = dH/dV = [H(liquid) - H(solid)]/[V(liquid) - V(solid)]

= [335,000 J/kg]/[1000⁻¹ - 915⁻¹ m³/kg]

= -3.61x10⁹ J/m³ = -3.61x10⁹ Pa

=> P₂ = P₁ - 3.61x10⁹ ln(T₂/T₁) Pa

or

T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa)

b) if the pressure in Denver is 84.6 kPa:

T₂(freezing) = 273.15exp[-(84,600-100,000)/(3.61x10⁹)]

≅ 273.15 = 0°C T₁(freezing) essentially no change

5 0
2 years ago
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