1 litre of water at 40°C is mixed with 1 litre of water at 60°C. What will be the approximate temperature of water after a certain time?
Between 40°C and 60°C
When two quantities of the same substance at different temperatures are mixed, heat energy is transferred from the hotter substance to the colder substance until they reach a thermal equilibrium, meaning they both reach the same final temperature. This process follows the principle of conservation of energy, often described by the principle of calorimetry.
The principle of calorimetry states that when heat exchange occurs between two or more bodies in an isolated system, the total heat lost by the hot bodies is equal to the total heat gained by the cold bodies.
In this specific case, we are mixing 1 litre of water at 40°C with 1 litre of water at 60°C. Both are the same substance (water) and have the same volume. Assuming the density and specific heat capacity of water are constant within this temperature range, equal volumes mean equal masses.
Let:
Since the volumes are equal and the substance is the same, we can assume \(m_1 = m_2 = m\).
Heat gained by the colder water = \(m_1 \times c \times (T_f - T_1)\)
Heat lost by the hotter water = \(m_2 \times c \times (T_2 - T_f)\)
According to the principle of calorimetry:
Heat gained = Heat lost
\(m \times c \times (T_f - T_1) = m \times c \times (T_2 - T_f)\)
Since \(m\) and \(c\) are the same and non-zero, we can cancel them:
\(T_f - T_1 = T_2 - T_f\)
\(2T_f = T_1 + T_2\)
\(T_f = \frac{T_1 + T_2}{2}\)
Substituting the given initial temperatures:
\(T_f = \frac{40°\text{C} + 60°\text{C}}{2}\)
\(T_f = \frac{100°\text{C}}{2}\)
\(T_f = 50°\text{C}\)
The approximate temperature of the water after mixing will be 50°C. This temperature is exactly halfway between the two initial temperatures, 40°C and 60°C.
Therefore, the approximate temperature of water after a certain time will be between 40°C and 60°C.
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