$解：水放出的熱量：$
$? Q_{吸}=c_{水}\ \mathrm {m}_{水}{?t}_{水}=4.2×{10}^3\ \mathrm {J/}(\mathrm {kg·℃})×0.5\ \mathrm {kg}×(60℃-40℃)=4.2×10^4\ \mathrm {J}?，$
$? Q_{吸}=Q_{放}?，$
$? ?根據(jù)?Q_{吸}=c_{酒精}\ \mathrm {m}_{酒精}{?t}_{酒精}?得：$
$? {?t}_{酒精}=\frac {Q_{吸}}{C_{酒精}\ \mathrm {m}_{酒精}}=\frac {4.2×{10}^4\ \mathrm {J}}{2.4×{10}^3\ \mathrm {J/}(\mathrm {kg·℃})×0.7\ \mathrm {kg}}=25℃?，$
$? t_{酒精}=t_{初}+{?t}_{酒精}=10℃+25℃=35℃.?$
