The temperature of 600 g of cold water rises by {{15}^{0}}, when 300 g of water {{50}^{0}} is added to it. What is the initial temperature of the cold water?A.{{10}^{0}}CB.{{15}^{0}}CC.{{5}^{0}}CD.{{25}^{0}}C

To solve this problem, we use the principle of conservation of energy, where the heat lost by the hot water equals the heat gained by the cold water.
Given:
• Mass of cold water (m1m_1) = 600 g
• Temperature rise of cold water (ΔT1\Delta T_1) = 15∘C15^\circ \mathrm{C}
• Mass of hot water (m2m_2) = 300 g
• Initial temperature of hot water (T2T_2) = 50∘C50^\circ \mathrm{C}
• Specific heat of water (cc) = constant (same for both waters, so it cancels out).
Let the initial temperature of the cold water be T1T_1.
Heat gained by cold water:
Qgain=m1c(T1+15−T1)=m1c(15)Q_{\text{gain}} = m_1 c (T_1 + 15 - T_1) = m_1 c (15)
Heat lost by hot water:
Qloss=m2c(T2−(T1+15))Q_{\text{loss}} = m_2 c (T_2 - (T_1 + 15))
Conservation of energy:
Qgain=QlossQ_{\text{gain}} = Q_{\text{loss}}
Substituting the expressions:
m1c(15)=m2c(50−(T1+15))m_1 c (15) = m_2 c (50 - (T_1 + 15))
Cancel cc (specific heat of water):
600(15)=300(50−T1−15)600 (15) = 300 (50 - T_1 - 15)
Simplify:
9000=300(35−T1)9000 = 300 (35 - T_1) 9000=10500−300T19000 = 10500 - 300 T_1 300T1=10500−9000300 T_1 = 10500 - 9000 300T1=1500300 T_1 = 1500 T1=1500300=5∘CT_1 = \frac{1500}{300} = 5^\circ \mathrm{C}
Final Answer:
The initial temperature of the cold water is 5∘C5^\circ \mathrm{C}.
Correct option: C) 5∘C5^\circ \mathrm{C}.