What Is The Heat Needed To Bring 120 G Of Ice From -170°C

What Is The Heat Needed To Bring 120 G Ice From 170c

Problem 1 (a) asks for the amount of heat required to warm 12.0 g of ice from -17.0°C to 0°C. The specific heat of ice is given as 2.00 kJ/kg°C. Problem 1 (b) requires calculations of the heat needed to convert 12.0 g of ice at 0°C into water at the same temperature, given the heat of fusion as 334 kJ/kg. Problem 1 (c) involves finding the heat to raise 12.0 g of water from 0°C to 25.0°C, with the specific heat of water being 4.19 kJ/kg°C. Problem 1 (d) asks for the volume of 12.0 g of ice using its density, 0.917 g/cm³, and examines whether the volume increases or decreases when ice converts to water, considering water's density as 1.00 g/cm³. Problem 1 (e) involves calculating the final temperature when 12.0 g of water at 25.0°C is mixed with 35.0 g of water initially at 167°F, assuming no heat loss and negligible container mass.

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The process of heating ice and water involves understanding multiple thermodynamic principles, including specific heat capacities, phase changes, density, and thermal equilibrium. The analysis begins with calculating the energy required to warm the ice, then proceeding to the phase transition, and finally considering temperature adjustments and volume changes upon phase conversion and mixing.

First, the energy needed to raise 12.0 grams of ice from -17.0°C to 0°C can be computed using the specific heat of ice. Since 1 kg of ice absorbs 2.00 kJ per degree Celsius, the small amount of 12.0 g (or 0.012 kg) necessitates converting mass to kilograms and applying the formula: Q = m c ΔT. Here, ΔT is 17.0°C (from -17.0°C to 0°C), leading to Q = 0.012 kg 2.00 kJ/kg°C 17.0°C, resulting in 0.408 kJ.

The next step involves calculating the heat required for melting the ice at 0°C. Using the heat of fusion of 334 kJ/kg, the energy needed is Q = m ΔH_fus, which gives 0.012 kg 334 kJ/kg = 4.008 kJ. This energy transition phase changes ice at melting point to water at the same temperature without temperature change.

Subsequently, raising 12.0 g of water from 0°C to 25.0°C involves the water's specific heat capacity. Converting the mass to kg, Q = 0.012 kg 4.19 kJ/kg°C 25.0°C yields approximately 1.255 kJ. Summing all these energies, the total heat required to bring the ice to 25°C water surpasses 5.668 kJ.

Volume calculations depend on the densities of ice and water. The volume of 12.0 g of ice is determined by dividing its mass by its density: V = m / ρ = 12.0 g / 0.917 g/cm³ ≈ 13.09 cm³. When ice melts into water, the volume change can be inferred from density differences: since water's density exceeds that of ice, the volume decreases upon melting, approximately by a factor of 0.917/1.00, leading to a reduction of about 8.3%.

The mixing of water at different temperatures involves heat transfer from the warmer to the cooler water until equilibrium is reached. Using the principle of conservation of energy, the heat lost by the warm water equals the heat gained by the cooler water. Setting up the equation: m₁ c (T_initial1 - T_final) = m₂ c (T_final - T_initial2), and solving for T_final, we find it to be approximately 74.7°C when m₁ = 12.0 g, T_initial1= 25.0°C, m₂= 35.0 g, and T_initial2= 167°F (which converts to about 75°C). The closer the initial temperatures, the quicker the thermal equilibrium is achieved.

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