Heat Transfer and Energy Balance: How Much Ice at 0°C Is Needed to Cool Water from 20°C to 10°C?
When dealing with thermal systems, the principle of conservation of energy is crucial for understanding the behavior of substances like water and ice. This article explores the problem of adding ice at 0°C to 1 kg of water initially at 20°C to cool it down to 10°C. We'll use the principle of conservation of energy to determine the mass of ice required.
Understanding the Problem
The problem involves a heat transfer process where heat is transferred from the water to the ice. The heat lost by the water must be equal to the heat gained by the ice. This balance is governed by the laws of thermodynamics. The goal is to calculate the mass of ice, ( m_i ), needed to achieve this cooling.
Parameters and Formulas
Mass of water (( m_w )): 1.0 kg Initial temperature of water (( T_{w_i} )): 20°C Final temperature of water (( T_f )): 10°C Initial temperature of ice (( T_i )): 0°C Specific heat of water (( c_w )): 4.18 kJ/kg°C Latent heat of fusion of ice (( L_f )): 334 kJ/kgCalculating Heat Transfer and Energy Balance
The heat lost by the water can be calculated using the formula:
[begin{align*} Q_w m_w cdot c_w cdot (T_{w_i} - T_f) 1.0 , text{kg} cdot 4.18 , text{kJ/kg°C} cdot (20°C - 10°C) 1.0 , text{kg} cdot 4.18 , text{kJ/kg°C} cdot 10°C 41.8 , text{kJ}end{align*} ]Next, we need to set up the energy balance for the ice. The ice first melts and then the resulting water warms up to the final temperature. The heat gained by the ice can be expressed as:
[begin{align*}Q_i m_i cdot L_f m_i cdot c_w cdot (T_f - T_i) m_i cdot 334 , text{kJ/kg} m_i cdot 4.18 , text{kJ/kg°C} cdot (10°C - 0°C) m_i cdot 334 , text{kJ/kg} m_i cdot 4.18 , text{kJ/kg°C} cdot 10°C m_i cdot 334 , text{kJ/kg} m_i cdot 41.8 , text{kJ/kg}end{align*} ]Combining these, we have:
[begin{align*}Q_i m_i cdot (334 , text{kJ/kg} 41.8 , text{kJ/kg}) m_i cdot 375.8 , text{kJ/kg}end{align*} ]To balance the heat transfer, we set (Q_w Q_i): [begin{align*}41.8 , text{kJ} m_i cdot 375.8 , text{kJ/kg} m_i frac{41.8 , text{kJ}}{375.8 , text{kJ/kg}} m_i approx 0.111 , text{kg}end{align*} ]
Conclusion
The mass of ice needed to cool 1 kg of water from 20°C to 10°C is approximately 0.111 kg or 111 grams. This calculation demonstrates the importance of understanding heat transfer and energy balance in practical applications such as refrigeration and air conditioning.