\(\displaystyle{s}={\frac{{{q}}}{{{m}\triangle{T}}}}\), so \(\displaystyle{q}={s}{x}{m}{x}\triangle{T}={\left({2.42}\ \frac{{J}}{{g}}-{K}\right)}{\left({62.0}\ {g}\right)}{\left({25.6}\ {K}\right)}={3.84}\times{10}^{{3}}\ {J}\)

(where \(\displaystyle\triangle{T}={40.8}{C}-{15.2}{C}={25.6}{C}\), but is equivalent to 25.6K as we are calculating the difference between the two temperatures \(\displaystyle{313.8}\ {K}-{288.2}\ {K}={25.6}\ {K}\))..

(where \(\displaystyle\triangle{T}={40.8}{C}-{15.2}{C}={25.6}{C}\), but is equivalent to 25.6K as we are calculating the difference between the two temperatures \(\displaystyle{313.8}\ {K}-{288.2}\ {K}={25.6}\ {K}\))..