in an enclosed container , 100g of ice at -10c and 200g of water at 25 c are mixed. Determine the amount of heat required to convert the mixture to steam at 110c
Cice=2100J/kg.K
Cwater=4186J/kg.K
Csteam=2080J/kg.K
And latent heats Lf ice=335000J/kg and Lv=2260000J/kg
To determine the amount of heat required to convert the mixture to steam at 110°C, we need to consider the heat required to warm up the ice and water, the heat required to melt the ice, and the heat required to convert the water to steam.
Let's break down the calculation step by step:
1. Heat required to warm up the ice from -10°C to 0°C:
- The specific heat capacity of ice (C_ice) given is 2100 J/kg.K.
- The mass of ice (m_ice) is 100g (or 0.1kg).
- The temperature change (ΔT) is 0°C - (-10°C) = 10°C.
- The heat required (Q_ice_warming) can be calculated using the formula:
Q_ice_warming = m_ice * C_ice * ΔT
2. Heat required to melt the ice at 0°C to water:
- The latent heat of fusion for ice (Lf_ice) given is 335000 J/kg.
- The heat required (Q_melting) can be calculated using the formula:
Q_melting = m_ice * Lf_ice
3. Heat required to warm up the water from 25°C to 100°C:
- The specific heat capacity of water (C_water) given is 4186 J/kg.K.
- The mass of water (m_water) is 200g (or 0.2kg).
- The temperature change (ΔT) is 100°C - 25°C = 75°C.
- The heat required (Q_water_warming) can be calculated using the formula:
Q_water_warming = m_water * C_water * ΔT
4. Heat required to convert the water to steam at 100°C to 110°C:
- The latent heat of vaporization for water (Lv) given is 2260000 J/kg.
- The mass of water (m_water) is 200g (or 0.2kg).
- The heat required (Q_vaporization) can be calculated using the formula:
Q_vaporization = m_water * Lv
5. Heat required to warm up the steam from 110°C to 100°C:
- The specific heat capacity of steam (C_steam) given is 2080 J/kg.K.
- The mass of steam (m_steam) can be determined by subtracting the total mass of ice and water from the initial mass of the mixture, which is 300g (or 0.3kg).
- The temperature change (ΔT) is 100°C - 110°C = -10°C (note the negative sign since it's cooling down).
- The heat required (Q_steam_warming) can be calculated using the formula:
Q_steam_warming = m_steam * C_steam * ΔT
The total heat required (Q_total) to convert the mixture to steam at 110°C can be obtained by summing up all the individual heat values calculated above:
Q_total = Q_ice_warming + Q_melting + Q_water_warming + Q_vaporization + Q_steam_warming
Substituting the given values and performing the respective calculations will yield the desired answer.