6 kg of cold water at 25°C is mixed with some boiled water at 100°C. The final temperature of the mixture is 60°C. How much boiled water was mixed?

• Mass of cold water, $m_c = 6 \text{ kg}$
• Initial temperature of cold water, $T_{ci} = 25^\circ\text{C}$
• Final temperature of the mixture, $T_f = 60^\circ\text{C}$
• Initial temperature of boiled water (hot water), $T_{hi} = 100^\circ\text{C}$ (assuming boiled water is at its boiling point).
• Specific heat capacity of water, $c$ (will cancel out).
• Mass of hot water, $m_h = ?$ (this is what we need to find).

Change in temperature for cold water:

Change in temperature for hot water:

Heat lost by hot water = Heat gained by cold water.
The formula for heat transfer is $Q = mc\Delta T$.

Since the specific heat capacity of water ($c$) is the same for both, it cancels out:

The calculated mass is $5.25 \text{ kg}$. Among the options, $5 \text{ kg}$ is the closest value.

The correct option is B) 5kg.

Last one for today! Hope it helped.