Answer:
1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.
Explanation:
Mass of ethylene glycol = m = 100 g
Specific heat capacity of ethylene glycol = c = 3.5 J/g°C
Change in temperature of ethylene glycol = ΔT
Heat loss by the ethylene glycol = Q = 350 J
ΔT = 1°C
1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.
Answer: ,
Explanation:
Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like and
These units of temperature are inter convertible.
We are given:
Temperature of the gas =
Converting this unit of temperature into by using conversion factor:
Converting this unit of temperature into by using conversion factor:
Thus the temperature on the Celsius and Kelvin scales are and
respectively.
Answer:
97 000 g Na
Explanation:
The absortion (or liberation) of energy in form of heat is expressed by:
q=m*Cp*ΔT
The information we have:
q=1.30MJ= 1.30*10^6 J
ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)
Cp=30.8 J/(K mol Na)
If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.
To do so, we use the molar mass of Na= 22.99g/mol
Now, we are able to solve for m:
=97 000 g Na