Q = mCΔT
Q is heat in joules, m is mass, C is specific heat, and delta T is change in temp
2099 J = (40.27g)(C)(148.5 - 24.8) = .421 J / gram K
<span>V = 24.0 mL + (35.2 g)(mL/10.5g) = I think i'm not all that sure but I think its this.</span>
Explanation:
Initial Pressure = 24 lb in-2
Initial Temperature = –5 o C = 268 K (Converting to kelvin temperature)
Final Pressure = ?
Final Temperature = 35 o C = 308 K (Converting to kelvin temperature)
No Change in Volume.
From Gay Lusaac's law; pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.
P1T1 = P2T2
P2 = P1T1 / T2
P2 = 24 * 268 / 308 = 20.88 lb in-2
There would be a drop in pressure as the temperature increases. Appropriate measures should b taken by regularly gauging the pressure of the tire.
Answer:
V2 = 35.967cm^3
Explanation:
Given data:
P1 = 0.2atm
P2 = 1.4atm
V1 = 250cm^3
V2 = ?
T1 = 10°C + 273 = 283K
T2 = 12°C + 273 = 285K
Apply combined law:
P1xV1/T1 = P2xV2/T2 ...eq1
Substituting values:
0.2 x 250/283 = 1.4 x V2/285
Solve for V2:
V2 = 14250/396.2
V2 = 35.967cm^3
