A 100 kg granite rock has been sitting in direct sunlight for 3 hours. It has absorbed 4.5x10^5 calories through its exposure to the sun. If it is now at a temp. of 50 Celsius, what was its temp. initially? (stone(granite) Specific heat in calories per gram degC is 0.19
sh in cal/kg deg C = .19 *1000 = 190
heat in = mass * sh * delta T
450,000 = 100 * 190 * (50-T)
23.68 = 50-T
T = 26.3 deg C
To find the initial temperature of the granite rock, we can use the formula for heat transfer:
Q = mcΔT
Where:
Q is the heat absorbed by the rock,
m is the mass of the rock,
c is the specific heat of the rock, and
ΔT is the change in temperature.
First, let's convert the mass of the granite rock from kilograms to grams:
100 kg = 100,000 grams
Next, we can rearrange the formula to solve for the initial temperature:
Q = mcΔT
ΔT = Q / mc
Given values:
Q = 4.5 x 10^5 calories
m = 100,000 grams
c = 0.19 calories/gram°C
Substituting the values into the equation:
ΔT = (4.5 x 10^5) / (100,000 x 0.19)
Now we can calculate ΔT:
ΔT = 4.5 x 10^5 / 19,000
ΔT = 23.6842°C
Finally, to find the initial temperature, we subtract ΔT from the final temperature:
Initial temperature = Final temperature - ΔT
Initial temperature = 50°C - 23.6842°C
Initial temperature ≈ 26.3158°C
Therefore, the initial temperature of the granite rock was approximately 26.3158°C.