Đáp án:
$a) \ t'_1=75,9^oC, m=0,143 \ kg$
$b) \ t=72,67^oC, t'=34,4^oC$
Giải:
`m_1=2 \ kg`
`t_1=80^oC`
`m_2=1 \ kg`
`t_2=20^oC`
Gọi `c` là nhiệt dung riêng của nước
a) Trong lần rót 1:
$t'_1=\dfrac{m_1ct_1+mct_2}{m_1c+mc}$
$t'_1=\dfrac{m_1t_1+mt_2}{m_1+m}$
$t'_1=\dfrac{2.80+m.20}{2+m}=\dfrac{160+20m}{m+2}$ (1)
Trong lần rót 2:
$t'_2=\dfrac{mct'_1+(m_2-m)ct_2}{mc+(m_2-m)}$
$t'_2=\dfrac{mt'_1+(m_2-m)t_2}{m+m_2-m}=\dfrac{mt'_1+m_2t_2-mt_2}{m_2}$
$28=\dfrac{mt'_1+1.20-m.20}{1}$
→ $mt'_1+20-20m=28$
→ $t'_1=\dfrac{8+20m}{m}$ (2)
Từ (1) và (2)
→ `\frac{160+20m}{m+2}=\frac{8+20m}{m}`
→ `160m+20m^2=20m^2+48m+16`
→ `112m=16`
→ `m=0,143 \ (kg)`
→ $t'_1=\dfrac{8+20.0,143}{0,143}=75,9 \ (^oC)$
b) Tương tự câu a):
$t=\dfrac{m_1t'_1+mt'_2}{m_1+m}=\dfrac{2.75,9+0,143.28}{2+0,144}=72,67 \ (^oC)$
$t'=\dfrac{mt+(m_2-m)t'_2}{m_2}=\dfrac{0,143.72,67+(1-0,143).28}{1}=34,4 \ (^oC)$
