$40$ g of hot water is poured in $100$ g of cold water, then temperature of cold water rises by $10^{o}$ C. If the temperature of hot water is $60^{o}$ C, calculate the initial temperature of cold water.

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Solution

Let the required final temperature be $T$ and initial temperature of cold water is $T_{c}$
Heat gain = Heat loss
$\Rightarrow 40 \times 1 \times (60 - T) = 100 \times 1 \times (T - T_{c}) \Rightarrow 40 \times 60 - 40 T = 100 \times 10 \Rightarrow 40 \times 6 - 4 T = 100 \Rightarrow 4 T = 140 ℃ \Rightarrow T = 35 ℃$

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40g of hot water is poured in 100g of cold water, then temperature of cold water rises by 10oC. If the temperature of hot water is 60oC, calculate the initial temperature of cold water.

Let the required final temperature be T and initial temperature of cold water is TcHeat gain = Heat loss⇒40×1×(60−T)=100×1×(T−Tc)⇒40×60−40T=100×10⇒40×6−4T=100⇒4T=140℃⇒T=35℃

$40$ g of hot water is poured in $100$ g of cold water, then temperature of cold water rises by $10^{o}$ C. If the temperature of hot water is $60^{o}$ C, calculate the initial temperature of cold water.