Answer:
The heat capacity for the sample is 0.913 J/°C
Explanation:
This is the formula for heat capacity that help us to solve this:
Q / (Final T° - Initial T°) = c . m
where m is mass and c, the specific heat of the substance
27.4 J / (80°C - 50°C) = c . 6.2 g
[27.4 J / (80°C - 50°C)] / 6.2 g = c
27.4 J / 30°C . 1/6.2g = c
0.147 J/g°C = c
Therefore, the heat capacity is 0.913 J/°C
Answer:
0.492 lb/in^3
Explanation:
From Density of Mercury (13.6g/cm3) given,
Mass = 13.6g
Volume = 1cm3
Now we covert these variables as follows :
For Mass:
454 g = 1 lb
13.6g = 13.6/454 = 0.03lb
For volume:
2.54 cm = 1 in
1cm = 1/2.54 = 0.394in
Volume = (0.394in)^3 = 0.061 in^3
Density = Mass /volume
Density = 0.03lb/0.061 in^3
Density = 0.492 lb/in^3
