How much heat is absorbed when a 35.0 g aluminum block increases in temperature from 23°C to 33°C?
Specific heat of Al: 0.900 J*g-1*°C-1
q=msΔT
q=(35.0 g)(.900)(33°C - 23°C)
q= 315 J? Just making sure there isn't anything that I am missing.
I didn't check the math on the calculator but the process and the numbers are correct.
Your calculation appears to be correct. To find the amount of heat absorbed, you used the formula q = msΔT, where q represents the heat absorbed, m is the mass of the aluminum block in grams, s is the specific heat of aluminum in J*g^(-1)*°C^(-1), and ΔT is the change in temperature.
In your calculation, you correctly plugged in the given values: m = 35.0 g, s = 0.900 J*g^(-1)*°C^(-1), and ΔT = (33°C - 23°C).
By substituting these values into the equation, you calculated:
q = (35.0 g)(0.900 J*g^(-1)*°C^(-1))(33°C - 23°C)
q = 315 J
Therefore, according to your calculation, the amount of heat absorbed when the 35.0 g aluminum block increases in temperature from 23°C to 33°C is 315 J.