anniN8yabrAlissym
contestada

#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 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.....................#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

Respuesta :

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.
obasola1

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.