A burn produced by live steam at 100oC is more severethan one produced by the same amount of water at 100oC.Verify this, calculate the heat that must be removed from 5.0 g ofwater at 100oC to lower its temperature to34oC.. Then calculate the heat that must beremoved from 5.0g of steam at 100oCto condense it andthen lower its temperature to 34oC, and compare thiswith the answer to part (a).
To determine whether a burn from live steam at 100°C is more severe than one from the same amount of water at 100°C, we need to understand the concept of specific heat capacity.
1. Specific Heat Capacity:
Specific heat capacity (c) is a measure of how much heat energy is required to raise the temperature of a substance by a certain amount. Different substances have different specific heat capacity values, and this affects how much heat energy they can absorb or release.
2. Calculation of Heat Energy:
The heat energy (Q) can be calculated using the equation: Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
a) Water at 100°C to 34°C:
Given:
Mass of water (m) = 5.0 g
Initial temperature (T1) = 100°C
Final temperature (T2) = 34°C
Specific heat capacity of water = 4.18 J/g°C (approximately)
Using the equation Q = mcΔT, we can calculate the heat energy as follows:
Q = (5.0 g) x (4.18 J/g°C) x (34°C - 100°C)
Q = (5.0 g) x (4.18 J/g°C) x (-66°C)
Q ≈ -13746 J (the negative sign indicates heat loss)
So, approximately 13,746 J of heat energy must be removed from 5.0 g of water at 100°C to lower its temperature to 34°C.
b) Steam at 100°C to 34°C (condensation and cooling):
To calculate the heat energy required to condense 5.0 g of steam at 100°C to water at 100°C, we need to consider the latent heat of vaporization (Lv) of water.
The value of Lv for water at its boiling point is approximately 2,260 J/g. This represents the amount of heat energy required to convert 1 gram of steam into 1 gram of water at the same temperature.
Given:
Mass of steam (m) = 5.0 g
Latent heat of vaporization (Lv) = 2,260 J/g (approximately)
The heat energy required for condensation is:
Q1 = (5.0 g) x (2,260 J/g)
Q1 = 11,300 J
To calculate the heat energy required to lower the temperature of the condensed water from 100°C to 34°C:
Q2 = (5.0 g) x (4.18 J/g°C) x (34°C - 100°C)
Q2 = (5.0 g) x (4.18 J/g°C) x (-66°C)
Q2 ≈ -13746 J (the negative sign indicates heat loss)
Thus, the total heat energy required for condensation and cooling is:
Q_total = Q1 + Q2
Q_total = 11,300 J + (-13,746 J)
Q_total ≈ -2,446 J (the negative sign indicates heat loss)
Comparing the results from part (a) and part (b), we observe that the heat energy required to lower the temperature from 100°C to 34°C is the same for both water and steam. However, the burn from live steam at 100°C is more severe due to the additional heat energy released during condensation.