Đáp án:

`m=0,07 \ kg`

Giải thích các bước giải:

`t_0=17^oC`

`t=37^oC`

`Q=6000 \ J`

$c=4200 \ J/kg.K$

The mass of hot water supplied by the shower in one second is:

`Q=mc(t-t_0)`

⇒ `m=\frac{Q}{c(t-t_0)}=\frac{6000}{4200.(37-17)}=0,07 \ (kg)`

Answer

$\\$ `Deltat^o = 37 - 17 = 20`

$\\$ `c = 4200J//kg.K`

$\\$ ______________________

$\\$ $\bullet$ The heat required to heat water to 37 degrees is :
$\\$ `Q = mcDeltat^o = m.4200. (37 - 17) = 84000m` 

$\\$ $\bullet$ The mas of hot water supplied by the shower in one second :
$\\$ `m = 6000/42000 = 1/14( kg)`