Question

$100g$ of ice is mixed with $200g$ of water at $20°C$. What is the temperature of the mixture?

Answer 1

Step 1: Given data

Mass of ice $=100g$

Mass of water $=200g$

Temperature $=20°C$

Specific heat of water $=1calorie$

Finding the temperature of the mixture

Step 2: Let us consider the temperature of the mixture be ${\theta }^{°}C$

$Heatgained=Heatlost$

$m_{1}C∆{\theta }_1=m_{2}C∆{\theta }_2$

$100(70-\theta )=200(\theta -20)$

$7000-100\theta =200\theta -4000$

$1100=300\theta$

$\theta =\frac{1100}{300}$

$\theta =3.6°C$

Therefore, the temperature of the mixture is $3.6°C$