I am lost about where to start on this.
"The idea comes to mind to put some hot rocks in your frozen bath water. What mass of rocks at 220 degrees Celsius needs to be added to 500.0L of frozen bath water at 0 degrees celsius to raise its temperature to 55 degrees celsius? The specific heat of the rcks is 2.5cal/g-C* and that of water is 1cal/g-C*. The heat of fusion for ice is 80cal/g. Assume there is no heat transfer between the ice/water and the walls of your bath.
Any help is GREATLY appreciated
It appears that ALL of the bath water is frozen and not just a part of it.
heat lost by rocks + heat absorbed by
ice melting + heat absorbed by liquid water to raise T to final T = 0
heat lost by rocks is
mass rocks x specific heat rocks x (Tfinal-Tinitial)
heat absorbed by ice to melt it.
mass ice x delta Hfusion
heat to raise temperature of liquid water.
mass water x specific heat H2O x (Tfinal-Tinitial)
Just add all of these in a string and solve for mass rocks. In a string it looks like this
[mass rocks x sp.h. rocks x (Tf-Ti)] + (mass ice x delta Hfusion) + [mass melted ice x sp.h H2O x (Tf-Ti)] = 0
Substitute and solve for mass rocks. The final answer is a large number.
THANK YOU!
To solve this problem, we need to use the principles of heat transfer and the specific heat equation. Here's how you can approach it step by step:
Step 1: Calculate the heat required to raise the temperature of the frozen bathwater from 0°C to 55°C.
- The specific heat equation is Q = m × c × ΔT, where Q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
- As the frozen bathwater is at 0°C and needs to reach 55°C, the change in temperature is ΔT = 55°C - 0°C = 55°C.
- The specific heat capacity of water is 1 cal/g-°C.
- The volume of the bathwater is given as 500.0 L. However, we need to convert this to mass in grams to use the specific heat equation. Use the density of water (1 g/mL) to convert: 500.0 L × 1 g/mL = 500.0 kg.
- Now we can calculate the heat required by substituting these values and solving the equation.
Q = m × c × ΔT
Q = 500.0 kg × 1 cal/g-°C × 55°C
Q = 27,500 cal.
Step 2: Calculate the heat released by the hot rocks.
- The hot rocks will cool down from 220°C to 55°C as they heat up the water.
- The specific heat capacity of the rocks is given as 2.5 cal/g-°C.
- We need to find the mass of the rocks required to release the same amount of heat absorbed by the water in Step 1.
- Rearrange the specific heat equation to solve for mass (m):
m = Q / (c × ΔT)
m = 27,500 cal / (2.5 cal/g-°C × (220°C - 55°C))
m = 27,500 cal / (2.5 cal/g-°C × 165°C)
m ≈ 66.7 g.
Therefore, you would need approximately 66.7 grams of hot rocks at 220°C to raise the temperature of 500.0 liters of frozen bathwater at 0°C to 55°C.