An object with a heat capacity of 3. 40×103j∘c absorbs 54. 0 kj of heat, beginning at −25. 0∘c. What will be the final temperature of the object?.

The final temperature of the object with a heat capacity of 3.40 × 10³ J °C absorbs 54.0 KJ of heat, beginning at −25.0 °C is - 9.12 °CC = q / ΔTΔT = T - ToC = Heat capacityq = Heat absorbed / releasedΔT = Change in temperatureT = Final temperatureTo = Initial temperatureC = 3.40 × 10³ J °C q = 54.0 KJ = 54 * 10³ JTo = - 25 °CT = ( q / C ) + ToT = ( 54 * 10³ / 3.40 × 10³ )  - 25T = 15.88 - 25T = - 9.12 °CIf heat is absorbed by a system, work is done on the system. So q will be taken as positive. If heat is released by a system, work is done by the system. So q will be taken as negative.  Therefore, the final temperature of the object is - 9.12 °CTo know more about Heat capacityhttps://brainly.com/question/1747943#SPJ4