An equal amount of water is taken in two different containers and their temperatures are $90^{o} C$ and $30^{o} C$ , respectively. If the rise in temperature of the cold water is equal to the fall in temperature of the hot water, then the common temperature of the mixture is ____

$45^{o} C$

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$66^{o} C$

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$30^{o} C$

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$60^{o} C$

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Solution

The correct option is D
$60^{o} C$

Step 1: Analyze the given situation and given data
The temperatures of the water in two containers are $90^{0} C$ and $30^{o} C$
Let the common temperature of the mixture be $T$ after it reaches equilibrium.
Step 2: Write expressions for heat lost and gained:
Heat lost by hot water will be $Q_{1} = m s (f i n a l T e m p - i n i t i a l T e m p_{1}) = m s (T - 90)$
Heat gained by cold water will be $Q_{2} = m s (f i n a l T e m p - i n i t i a l T e m p_{2}) = m s (T - 30)$
Step 3: Finding the common temperature
$m$ and $s$ will remain the same because an equal quantity of water is taken
By the principle of calorimetry, $Q_{1} = Q_{2}$ , we have,
$m s (T - 90) = m s (30 - T)$
or, $T - 90 = 30 - T$
or, $2 T = 120$
or, $T = 60^{o} C$
So, the correct option is (D) $60^{o} C$ .

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An equal amount of water is taken in two different containers and their temperatures are 90oC and 30oC, respectively. If the rise in temperature of the cold water is equal to the fall in temperature of the hot water, then the common temperature of the mixture is ____

An equal amount of water is taken in two different containers and their temperatures are $90^{o} C$ and $30^{o} C$ , respectively. If the rise in temperature of the cold water is equal to the fall in temperature of the hot water, then the common temperature of the mixture is ____