The temperature of 600 g of cold water rises by 15o C when 300 g of hot water at 50-C is added to it- What was the initial temperature of the cold water-

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Question

The temperature of $600$ g of cold water rises by $15^{o} C$ when $300$ g of hot water at $50^{\circ} C$ is added to it. What was the initial temperature of the cold water?

{2.5}^{\circ} C

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5^{\circ} C

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{7.5}^{\circ} C

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10^{\circ} C

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Solution

The correct option is B $5^{\circ} C$
Let initial temperature be $t_{1}$ .
The final temperature is given as $t_{1} + 15 = \frac{m_{1} t_{1} + m_{2} t_{2}}{m_{1} + m_{2}} = \frac{600 \times t_{1} + 300 \times 50}{600 + 300} = (600 t_{1} + 15000) / 900$
$\Rightarrow 300 t_{1} = 1500$
$\Rightarrow t_{1} = 5^{o} C$

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The temperature of 600 g of cold water rises by 15oC when 300 g of hot water at 50∘C is added to it. What was the initial temperature of the cold water?

The correct option is B 5∘CLet initial temperature be t1.The final temperature is given as t1+15=m1t1+m2t2m1+m2=600×t1+300×50600+300=(600t1+15000)/900⇒300t1=1500⇒t1=5oC

The temperature of $600$ g of cold water rises by $15^{o} C$ when $300$ g of hot water at $50^{\circ} C$ is added to it. What was the initial temperature of the cold water?