Answer:
2621.75 j heat is required to increase the temperature 25.5°C to 46°C.
Explanation:
Given data:
Mass of sample = 142.1 g
Initial temperature = 25.5°C
Final temperature = 46°C
Specific heat capacity of Al = 0.90 J/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 46°C - 25.5°C
ΔT = 20.5°C
Q = 142.1 × 0.90 J/g.°C × 20.5°C
Q = 2621.75 j
Thus, 2621.75 j heat is required to increase the temperature 25.5°C to 46°C.
Answer:
2.99 x 10^3 s
Explanation:
= 2.64 + .347 X 10^3
= 2.987 X 10 ^3 = 2.99 X 10^3 With 3 signif digits
Answer:
Normal weather is like rn with no hurricanes or anything dangerous
Severe weather, is a weather that is that can cause a lot of harm.
Well a compound is 2 or more different molecules together. So if you see a periodic table you will see the elements to see if they are the same or not.
Answer:
Q = 43,000 calories = 180,000 joules = 180 Kilojoules
Explanation:
Heat flow for phase change can be defined as Q = m·ΔHv
ΔHv = Heat of Vaporization for water = 540 calories/gram = 2259 joules/gram = 2.259 Kilojoules/gram
m = mass of object of interest ( in this problem 80 grams steam 100°C converting to 80 grams of water at 100°C
Q = m·ΔHv = 80g x 540 cals/g =43.200 calories x 4.184 joules/cal = 180,748.8 joules (calculator answer) => 180,000 joules (2 sig. figs.) = 180 Kj