For this problem, we use the formula for sensible heat which is written below:
Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference
Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
<span>The wave nature of an electron is indicated by its motion and the diffraction and interference of electrons in a beam.</span>
Answer:
a ) 1876.14 grams CaBr2
b ) 19.78 grams N2
sorry..i only have time to do the first two :)
Explanation:
Answer:
31.1°C
Explanation:
Given parameters:
Temperature = 88°F
The formula of the to convert is:
T°F = T°C - 32 / 1.8 = 
Now input the parameters and solve;
T°F =
T°F = 31.1°C
Given:
Q = 9.4 kJ/(kg-h), the heat production rate
c = 4.18 J/(g-K), the heat capacity
t = 2.5 h, amount of time
Note that
c = 4.18 J/(g-K) = 4180 J/(kg-K) = 4.18 kJ/kg-K)
Consider 1 kg of mass.
Then
Qt = cΔT
where ΔT is the increase in temperature (°K)
(1 kg)*(9.4 kJ/(kg-h))*(2.5 h) = (1 kg)*(4.18 kJ/(kg-K))*(ΔT K)
23.5 = 4.18 ΔT
ΔT = 23.5/4.18 = 5.622 K = 5.622 °C
Answer: 5.62 K (or 5.62 °C)
