You’re answer is C Explaination if the pit side was facing earth it wouldn’t be a new moon
Answer:
you are rigth
Explanation:
for the bottom you did extra credit
Answer:
a. 50KCal
b. 400KCal
c. Same as (a) above
Explanation:
Given
To raise the temperature of 1kg of liquid water at 1°C requires 1KCal
To raise the temperature of 1kg of ice or water vapour by 1°C requires 0.5KCal
To melt 1kg of ice at 0°C requires 80KCal
To evaporate 1kg of liquid water sitting at 100°C requires 540KCal
a. How much heat is required to raise the temperature of 5 kg of liquid water by 20 C?
To raise the temperature of 5 kg of ice by 20°C requires:
5 kg * (0.5 kcal / kgC) * 20C
= 50 KCal
b. How much heat is required to melt 5 kg of ice at 0 C?
To melt an ice of 5 kg of ice at 0 C requires:
5 kg * (80 kcal / kg)
= 400 KCal
c. Same as (a) above
Heat = mass * heat capacity of water * change in temperature
mass = 5.25 g
heat capacity of water = 4.186 joule/gram °C
Change in temperature = 62.8°C - 5.3°C = 57.5 °C
Plug in the values
heat = 5.25 g * 4.186 joule/gram °C * 57.5 °C = 1263.6 J
Rounded to two three significant figures, it is 1260 J of energy needed.
In terms of calories, the heat capacity of water is 1 calorie/gram °C. So do the plugging in all over again.
mass = 5.25 g
heat capacity of water = 1 calorie/gram °C
Change in temperature = 62.8°C - 5.3°C = 57.5 °C
heat = 5.25 g * 1 calorie/gram °C * 57.5 °C = 301.9 calories
Rounded to 3 significant figures, it is 302 calories
Q=SM∆T=4.18*5.25*(62.8-4.3)=1280 J
1280 J * (1 cal/4.18 J) = 307 cal
