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Question
Calculate the mass of ice required to lower the temperature of 300 g of water 40°C to water at 0°C.
(Specific latent heat of ice = 336 J g−1, the Specific heat capacity of water = 4.2J g−1 °C−1)
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Solution
Given: Mass of water, \(m_1=300\,\text{g}\)
Initial temperature of water, \(T_1=40^\circ\text{C}\)
Final temperature, \(T_2=0^\circ\text{C}\)
Specific heat capacity of water, \(c=4.2\,\text{J g}^{-1}\,^\circ\text{C}^{-1}\)
Specific latent heat of fusion of ice, \(L=336\,\text{J g}^{-1}\)
Assuming the ice is initially at 0°C and no heat is lost to the surroundings,
According to the principle of calorimetry,
\[ \text{Heat lost by water}=\text{Heat gained by ice} \]
The heat lost by water.
\[ Q=m_1c(T_1-T_2) \]
\[ =300\times4.2\times(40-0) \]
\[ =50400\,\text{J} \]
The mass of ice required.
Heat gained by ice during melting:
\[ Q=m_2L \]
\[ 50400=m_2\times336 \]
\[ m_2=\frac{50400}{336} \]
\[ {m_2=150\,\text{g}} \]
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| Column A | Column B | ||
| 1. | Specific heat capacity of water | a. | 0°C |
| 2. | Latent heat of fusion of ice | b. | 2260 J/g |
| 3. | Latent heat of vaporization of water | c. | 100°C |
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