Question

You have 750 g of water at 10.0 C in a large insulated beaker. How much boiling water at 100.0 degrees C must you add to this beaker so that the final temperature of the mixture will be 75 degrees C?

Answer 1

Here the concept is that

loss of heat by hot body \(=\) gain of heat by cold body

\(\mathrm{m}_{2} \mathrm{c}\left(\mathrm{T}_{2}-\mathrm{T}_{\mathrm{f}}\right)=\mathrm{m}_{1} \mathrm{c}\left(\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{1}\right)\)

here \(\mathrm{m}_{2}\) is the mass of the hot water.

c is the specific heat of water \(\mathrm{T}_{2}\) is the hot body temperature \(=100^{\circ} \mathrm{C}\)

\(\mathrm{T}_{\mathrm{f}}\) is the final temperature of the mixture \(=75^{\circ} \mathrm{C}\) \(\mathrm{T}_{1}\) is the cold temperature \(=10^{\circ} \mathrm{C}\) \(\mathrm{m}_{1}\) is the mass of the cold water \(=750 \mathrm{~g}=0.75 \mathrm{~kg}\) Hence \(\left(\mathrm{m}_{2}\right)(\mathrm{c})(100-75)=(0.75 \mathrm{~kg})(\mathrm{c})(75-10)\)

\(\Rightarrow \mathrm{m}_{2}(25)=(0.75 \mathrm{~kg})(65)\)

\(\Rightarrow \mathrm{m}_{2}=\frac{0.75 \mathrm{~kg}(65)}{25 \mathrm{~kg}}=1.95 \mathrm{~kg}\)