ThermodynamicsHard
Calorimetry with melting ice
0.10 kg of ice at 0 °C is dropped into 0.50 kg of water at 25 °C (adiabatic container). Find the equilibrium temperature. (λ = 334000 J/kg, c = 4186 J/kg·K)
Given data
m_ice = 0.10 kg ice (0 °C)m_w = 0.50 kg water (25 °C)λ = 334000 J/kgc = 4186 J/kg·K- The ice is already at 0 °C: before it can warm up it must melt. Melting absorbs latent heat at constant temperature (the energy breaks the crystal lattice, it does not raise T). Compute the heat needed to melt it all: Q_f = m_ice·λ.
- Check whether the ice really melts completely. Compare Q_f with the maximum heat the water can release cooling down to 0 °C: Q_w = m_w·c·ΔT. If Q_w > Q_f, all the ice melts and energy is left to warm everything above 0 °C.
- The container is adiabatic: no heat escapes. So the heat released by the water = heat absorbed by the ice (to melt it + to warm the meltwater from 0 °C to T). Impose this balance and solve for the equilibrium temperature: m_w·c·(25−T) = m_ice·λ + m_ice·c·T.
Full worked solution
- The ice is already at 0 °C: before it can warm up it must melt. Melting absorbs latent heat at constant temperature (the energy breaks the crystal lattice, it does not raise T). Compute the heat needed to melt it all: Q_f = m_ice·λ..
- Check whether the ice really melts completely. Compare Q_f with the maximum heat the water can release cooling down to 0 °C: Q_w = m_w·c·ΔT. If Q_w > Q_f, all the ice melts and energy is left to warm everything above 0 °C.: all the ice melts.
- The container is adiabatic: no heat escapes. So the heat released by the water = heat absorbed by the ice (to melt it + to warm the meltwater from 0 °C to T). Impose this balance and solve for the equilibrium temperature: m_w·c·(25−T) = m_ice·λ + m_ice·c·T..
Result: °C