<prompt> Theha equation ea lipalo ho hlalosa motsamao oa 100 g ea ntho ho tloha mocheso oa 273 K ho ea ho 293 K.nan  <response>: Instruction in English: Formulate a mathematical equation to express the movement of 100 g of a substance from a temperature of 273 K to a temperature of 293 K. . Response in English: q m × c × ΔTwhere:- q joules or J is the amount of heat absorbed or released,- m grams or g is the mass of the substance,- c joule per gram kelvin or J g-1 K-1 is the specific heat capacity of the substance, and- ΔT kelvin or K is the change in temperature of the substance that is, 293 K - 273 K 20 K.So, for 100 g of a substance moving from a temperature of 273 K to a temperature of 293 K, the equation would be:q 100 g × c × 20 Kor q 2000 × c JNote that the specific heat capacity c, which varies for different materials and is temperature dependent, needs to be known, or measured on calorimeter, in order to calculate the exact heat absorbed or released. . Response in Sesotho: q m × c × ΔT moo: - q joules kapa J ke boholo ba mocheso o monyetsoeng kapa o lokolloang, - m ligrama kapa g ke boima ba ntho, - c joule ka gram kelvin kapa J g -1 K-1 ke matla a khethehileng a mocheso oa ntho, me - ΔT kelvin kapa K ke phetoho ea mocheso oa ntho ke hore, 293 K - 273 K 20 K. Kahoo, bakeng sa 100 g ea ntho e tlohang mochesong oa 273 K ho ea ho 293 K, equation e tla ba: q 100 g × c × 20 K kapa q 2000 × c J Hlokomela hore matla a khethehileng a mocheso c, e fapaneng bakeng sa lisebelisoa tse fapaneng me e itšetlehile ka mocheso, e hloka ho tsejoa, kapa ho lekanyetsoa ka calorimeter, e le ho bala mocheso o tobileng o monyelitsoeng kapa o lokollotsoeng. 