Answer:
1.23 j/g. °C
Explanation:
Given data:
Mass of metal = 35.0 g
Initial temperature = 21 °C
Final temperature = 52°C
Amount of heat absorbed = 320 cal (320 ×4.184 = 1338.88 j)
Specific heat capacity of metal = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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 = 52°C - 21 °C
ΔT = 31°C
1338.88 j= 35 g ×c× 31°C
1338.88 j= 1085 g.°C ×c
1338.88 j/1085 g.°C = c
1.23 j/g. °C = c
Answer:
D.
Explanation:
I had looked it up on the quiz site
Answer:
5.6
Explanation:
Because of the gravity of the earth
<u>Answer:</u>
<em>When we finish, the temperature would be 32.5℃</em>
<em></em>
<u>Explanation:</u>
Density of water = mass/volume
So,
Mass of water = Density × Volume
where
= Final T - Initial T
Q is the heat energy in calories
c is the specific heat capacity (for water 1.0 cal/(g℃))
m is the mass of water
plugging in the values
Final T = ∆T + Initial T
= 7.5℃ + 25℃ = 32.5℃ (Answer).
<span>Colligative properties are properties of solutions that depend on the number of molecules [or ions] in a given volume of solvent and not on the properties (e.g. size or mass) of the compound. Colligative properties include: lowering of vapor pressure; elevation of boiling point; depression of freezing point and osmotic pressure.</span>
