The SI unit of specific heat is J per gram per degree
Celsius. Thus it follows that specific heat could be calculated in this way:
Specific Heat = Energy / (mass x change in temperature)
Thus,
Specific Heat = 3.912 cal / (9.84 oz x (191.2 ˚F – 73.2 ˚F))
Specific Heat = 3.369 x 10^-3 cal/oz-˚F
Answer:
Explanation:
we know that specific heat is the amount of heat required to raise the temperature of substance by one degree mathmeticaly
Q=mcΔT
ΔT=T2-T1
ΔT=26.8-10.2=16.6
C for water is 4.184
therefore
Q=1.00*4.184*16.6
Q=69.4 j
now we have to covert joule into calorie
1 calorie =4.2 j
x calorie=69.4 j/2
so 69.4 j =34.7 calorie thats why 34.7 calorie heat is required to raise the temperature of water from 10.2 to 26.8 degree celsius
The question is incomplete , the complete question is;
The mass of each Wt isotope is measured relative to C-12 and tabulated below. Use the mass of C-12 to convert each of the masses to amu and calculate the atomic mass of Wt.
Wt-296 = 24.6622
Answer:
The atomic mass of Wt-296 is 195.9464 amu.
Explanation:
1 amu is defined as 1 by twelfth of the carbon-12 mass.
Mass of an isotope Wt-296 = 24.6622
Mass of an isotope Wt-296 in amu = M
The atomic mass of Wt-296 is 195.9464 amu.
