Question

A copper vessel of mass 100 g contains 150 g of water at 50°C. How much ice is needed to cool it to 5°C?

Given: Specific heat capacity of copper = 0.4 Jg⁻¹ °C⁻¹

The Specific heat capacity of water = 4.2 Jg⁻¹ °C⁻¹

The Specific latent heat of fusion ice = 336 Jg⁻¹

Answer 1

Given: Mass of copper vessel, \(m_1=100\,\text{g}\)

Mass of water, \(m_2=150\,\text{g}\)

Initial temperature, \(T_1=50^\circ\text{C}\)

Final temperature, \(T_2=5^\circ\text{C}\)

Specific heat capacity of copper, \(c_1=0.4\,\text{J g}^{-1}\,^\circ\text{C}^{-1}\)

Specific heat capacity of water, \(c_2=4.2\,\text{J g}^{-1}\,^\circ\text{C}^{-1}\)

Specific latent heat of fusion of ice, \(L=336\,\text{J g}^{-1}\)

Assumption: Ice is initially at 0°C, and no heat is lost to the surroundings.

Now, heat energy lost by the copper vessel is:

\[ Q_1=m_1c_1(T_1-T_2) \]

\[ =100\times0.4\times(50-5) \]

\[ =1800\,\text{J} \]

Similarly, heat energy lost by water is:

\[ Q_2=m_2c_2(T_1-T_2) \]

\[ =150\times4.2\times(50-5) \]

\[ =28350\,\text{J} \]

Hence, the total heat energy lost is:

\[ Q=Q_1+Q_2 \]

\[ =1800+28350=30150\,\text{J} \]

Ice at 0°C melts and the resulting water warms to 5°C.

\[ Q=mL+mc_2(5-0) \]

\[ =336m+4.2m\times5 \]

\[ =357m \]

According to the principle of calorimetry,

\[ \text{Heat lost}=\text{Heat gained} \]

\[ 30150=357m \]

\[ m=\frac{30150}{357}\]

\[=84.45\,\text{g} \]