Question

hot water is added to three times its mass of water at 10 degree celsius and the resulting temperature is 20 degrees Celsius What is the initial temperature of the hot water​

Answer 1

Answer:

The initial temperature of the hot water is $50\; \rm ^{\circ} C$ (assuming that no heat was lost to the surroundings.)

Explanation:

Let $m$ denote the mass of the hot water.

The question states that the mass of the water at $10\; \rm ^\circ C$ is three times the mass of the hot water. If the mass of the hot water is $m$, the mass of the cold water would be $3\, m$.

Let $c$ denote the specific heat capacity of water. Let $m$ denote the mass of some water. The energy required to change the temperature of that much water by $\Delta T$ (without state change) would be:

$Q = c \cdot m \cdot \Delta T$.

The temperature change for the cold water was:

$\Delta T_1 = 20\; \rm ^{\circ} C - 10\; \rm ^{\circ} C = 10\; \rm K$.

Energy required to raise the temperature of water with mass $3\, m$ from $10\; \rm ^{\circ} C$ to $20\; \rm ^{\circ} C$:

$Q_1 = c \cdot (3\, m) \cdot (10\; \rm K)$.

On the other hand, if the initial temperature of the hot water is $t\; \rm ^{\circ} C$ (where $t > 20$,) the temperature change would be:

$\Delta T_2 = t\; {\rm ^{\circ} C} - 20\; {\rm ^{\circ} C} = (t - 20)\; {\rm K}$.

Calculate the energy change involved:

$Q_2 = c \cdot m \cdot ((t - 20)\; \rm K)$.

If no energy was lost to the surroundings, $Q_1$ should be equal to $Q_2$. That is:

$c \cdot (3\, m) \cdot (10\; {\rm K}) = c\cdot m \cdot ((t - 20)\; {\rm K})$.

Simplify and solve for $t$:

$t - 20 = 30$.

$t = 50$.

Therefore, the initial temperature of the hot water would be $50\; {\rm ^\circ C}$.