Given 16.2 grams of substance Y, if the substance absorbs 2722 joules of energy and the specific heat of the substance is 9.22 J/g·°C, what is the final temperature of the substance if the initial was 26 degrees Celsius?

Final temperature (T) = 44.22°C (Approx)

Explanation:

Given:

Mass of Y (m) = 16.2 gram

Energy required (q) = 2722 J

Specific heat of the substance(c) = 9.22 J/g·°C

Change in temperature ΔT = T - 26°C

Find:

Final temperature (T)

Computation:

q = mcΔT

2722 = (16.2)(9.22)(T - 26°C)

18.224 = T - 26°C

T = 26 + 18.224

T = 44.224°C

Final temperature (T) = 44.22°C (Approx)

oh it is true if thats what you are asking

the temperature of the gas, in k, not celsius.