Respuesta :
Answer:
The initial temperature of the hot water is [tex]50\; \rm ^{\circ} C[/tex] (assuming that no heat was lost to the surroundings.)
Explanation:
Let [tex]m[/tex] denote the mass of the hot water.
The question states that the mass of the water at [tex]10\; \rm ^\circ C[/tex] is three times the mass of the hot water. If the mass of the hot water is [tex]m[/tex], the mass of the cold water would be [tex]3\, m[/tex].
Let [tex]c[/tex] denote the specific heat capacity of water. Let [tex]m[/tex] denote the mass of some water. The energy required to change the temperature of that much water by [tex]\Delta T[/tex] (without state change) would be:
[tex]Q = c \cdot m \cdot \Delta T[/tex].
The temperature change for the cold water was:
[tex]\Delta T_1 = 20\; \rm ^{\circ} C - 10\; \rm ^{\circ} C = 10\; \rm K[/tex].
Energy required to raise the temperature of water with mass [tex]3\, m[/tex] from [tex]10\; \rm ^{\circ} C[/tex] to [tex]20\; \rm ^{\circ} C[/tex]:
[tex]Q_1 = c \cdot (3\, m) \cdot (10\; \rm K)[/tex].
On the other hand, if the initial temperature of the hot water is [tex]t\; \rm ^{\circ} C[/tex] (where [tex]t > 20[/tex],) the temperature change would be:
[tex]\Delta T_2 = t\; {\rm ^{\circ} C} - 20\; {\rm ^{\circ} C} = (t - 20)\; {\rm K}[/tex].
Calculate the energy change involved:
[tex]Q_2 = c \cdot m \cdot ((t - 20)\; \rm K)[/tex].
If no energy was lost to the surroundings, [tex]Q_1[/tex] should be equal to [tex]Q_2[/tex]. That is:
[tex]c \cdot (3\, m) \cdot (10\; {\rm K}) = c\cdot m \cdot ((t - 20)\; {\rm K})[/tex].
Simplify and solve for [tex]t[/tex]:
[tex]t - 20 = 30[/tex].
[tex]t = 50[/tex].
Therefore, the initial temperature of the hot water would be [tex]50\; {\rm ^\circ C}[/tex].