A 34.0 g piece of metal is heated to 92.0°C then placed in a beaker of water containing 22.0 g of water at 19.0°C. The temperature of the water rises to 24.0°C. What is the specific heat of the metal?
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Ответ:
0.1988 J/g°C
Explanation:
-Qmetal = Qwater
Q = mc∆T
Where;
Q = amount of heat
m = mass of substance
c = specific heat of substance
∆T = change in temperature
Hence;
-{mc∆T} of metal = {mc∆T} of water
From the information provided in this question, For water; m= 22.0g, ∆T = (24°C-19°C), c = 4.18J/g°C.
For metal; m= 34.0g, ∆T = (24°C-92°C), c = ?
Note that, the final temperature of water and the metal = 24°C
-{34 × c × (24°C-92°C)} = 22 × 4.18 × (24°C-19°C)
-{34 × c × (-68°C)} = 459.8
-{34 × c × -68} = 459.8
-{-2312c} = 459.8
+2312c = 459.8
c = 459.8/2312
c = 0.1988
The specific heat capacity of the metal is 0.1988 J/g°C
Ответ:
14 mol
Explanation:
CH4(g) + 2O2(g) » CO2(g) + 2H2O(g)
from reaction 1 mol 2 mol
given 7 mol x mol
x = (7*2)/1 = 14 mol