Water can be heated in tightly woven baskets by dropping in hot rocks. The technique is useful for cooking in the absence of metal or pottery vessels. Estimate the size (in liters) of basket needed to hold the rocks and water if 5 L of boiling-hot (100°C) water is required. The water comes from a well at 10°C, and the rocks (granite with a density of 2.7 g cm−3) come from a hot fire at 900°C. Take data from Table 10−1. List the assumptions required to make the estimate.
A)6 L
Assumptions:
(1) the density of the water does not change between 10.0°C to 100°C
(2) the density of the rock is 2.7 g cm−3 at 100°C
(3) the energy created by the rocks are transfered to the water at a constant rate.
(4) heat is lost to the surroundings through the walls of the basket
(5) The rocks decrease in size as a function of temperature
B)6 L
Assumptions:
(1) the density of the water does not change between 10.0°C to 100°C
(2) the density of the rock is 2.7 g cm−3 at 100°C
(3) the specific heat capacities of the water and rock do not change with temperature
(4) no heat is lost to the surroundings through the walls of the basket
(5) no heat is used to warm up the walls of the basket
C)6 L
Assumptions:
(1) the density of the water does not change between 10.0°C to 100°C
(2) the density of the rock is 2.7 g cm−3 at 100°C
(3) the energy created by the rocks are transfered to the water at a constant rate.
(4) no heat is lost to the surroundings through the walls of the basket
(5) no heat is used to warm up the walls of the basket
D)8 L
Assumptions:
(1) the density of the water does not change between 10.0°C to 100°C
(2) the density of the rock is 2.7 g cm−3 at 100°C
(3) the specific heat capacities of the water and rock do not change with temperature
(4) no heat is lost to the surroundings through the walls of the basket
(5) no heat is used to warm up the walls of the basket
E)8 L
Assumptions:
(1) the density of the water does not change between 10.0°C to 100°C
(2) the density of the rock is 2.7 g cm−3 at 100°C
(3) the specific heat capacities of the water and rock do not change with temperature
(4) heat is lost to the surroundings through the walls of the basket
(5) The rocks decrease in size as a function of temperature
F) 8 L
Assumptions:
(1) the density of the water does not change between 10.0°C to 100°C
(2) the density of the rock is 2.7 g cm−3 at 100°C
(3) the energy created by the rocks are transfered to the water at a constant rate.
(4) heat is lost to the surroundings through the walls of the basket
(5) The rocks decrease in size as a function of temperature
The correct answer is option E) 8 L.
To estimate the size of the basket needed to hold both the rocks and water, we need to consider several assumptions and calculations.
Assumptions:
1) The density of the water does not change between 10.0°C to 100°C. This assumption is reasonable as the change in density of water over this temperature range is negligible.
2) The density of the rock is 2.7 g cm−3 at 100°C. This assumption is based on the provided data (Table 10-1).
3) The specific heat capacities of the water and rock do not change with temperature. This assumption is reasonable for the temperature range mentioned.
4) Heat is lost to the surroundings through the walls of the basket. Heat loss due to conduction and convection can occur, affecting the efficiency of the heating process.
5) The rocks decrease in size as a function of temperature. This assumption is reasonable as most materials contract with increasing temperature.
To estimate the size of the basket, we can use the principle of conservation of energy.
First, we need to calculate the amount of heat required to raise the temperature of 5 L of water from 10°C to 100°C using the equation:
Q = m * c * ΔT
Where:
Q = Heat energy (in Joules)
m = Mass of water (in kg)
c = Specific heat capacity of water (in J/(kg·K))
ΔT = Change in temperature (in K)
The density of water is approximately 1 g/cm³, so the mass of 5 L of water can be calculated as:
mass = volume * density = 5000 cm³ * 1 g/cm³ = 5000 g = 5 kg
Now, using the specific heat capacity of water (4.184 J/(g·K)), we can calculate the heat required:
Q_water = 5 kg * 4.184 J/(g·K) * (100°C - 10°C) = 2084 J/g * (90 K) = 187,560 J
Next, we need to calculate the amount of heat released by the rocks as they cool down from 900°C to a temperature close to the boiling point of water (100°C). The heat released can be calculated using the equation:
Q_rock = m * c * ΔT
Where:
m = Mass of rocks (in kg)
c = Specific heat capacity of rocks (in J/(kg·K))
ΔT = Change in temperature (in K)
First, we need to find the mass of the rocks using their density of 2.7 g/cm³. Let's assume the mass of the rocks is "m" kg.
density_rock = 2.7 g/cm³ = m / V
2.7 g/cm³ = m / volume_of_rocks
2.7 g/cm³ = m / (volume_of_rocks_water + volume_of_rocks)
2.7 g/cm³ = m / (5000 cm³ + volume_of_rocks)
We know that the volume of rocks is equal to the volume of water displaced by the rocks. Assuming the rocks displace an equal volume, we have:
volume_of_rocks = 5000 cm³
Substituting this into the equation:
2.7 g/cm³ = m / (5000 cm³ + 5000 cm³)
2.7 g/cm³ = m / 10000 cm³
2.7 g/cm³ * 10000 cm³ = m
27,000 g = m = 27 kg
Now, using the specific heat capacity of rocks (0.84 J/(g·K)), we can calculate the heat released:
Q_rock = 27 kg * 0.84 J/(g·K) * (900°C - 100°C) = 22,680 J/g * (800 K) = 18,144,000 J
The total heat required to raise the temperature of the water and the total heat released by the rocks should be equal (assuming no other energy losses), so we can set up the equation:
Q_water = Q_rock
187,560 J = 18,144,000 J, which is clearly not equal.
Therefore, we can conclude that the assumption that no heat is used to warm up the walls of the basket (option E) 8 L) is incorrect. Heat is indeed transferred to the walls of the basket, leading to additional heat loss and a larger size requirement for the basket.
Hence, the correct answer is option D) 8 L, where the assumption is that heat is lost to the surroundings through the walls of the basket.