QQuestionPhysics
QuestionPhysics
The specific heat of water is 1.0 Btu/lb°F, and its density is 62.4 lb/ft³. Determine the specific heat capacity of water.
10 months agoReport content
Answer
Full Solution Locked
Sign in to view the complete step-by-step solution and unlock all study resources.
Step 1:: Understand the problem
The specific heat of a substance is the amount of heat energy required to raise the temperature of one unit mass by one unit of temperature. In this case, we are given the specific heat of water in British Thermal Units (Btu) per pound-mass and degree Fahrenheit. Our goal is to find the specific heat capacity, which is the specific heat multiplied by the mass.
Step 2:: Gather necessary information
- Density of water ($$\rho_{water}$$): 62.4 lb/ft³
We are given: To find the specific heat capacity, we need to convert the density from pounds per cubic foot to pounds per unit of volume (e.g., pound, gram, etc.). Since the specific heat is already given in the correct units, we don't need to convert it.
Step 3:: Convert density units
\rho_{water} = 62.4 \frac{lb}{ft^3} = 62.4 \frac{lb}{1728 in^3} \approx 0.036 \frac{lb}{in^3}
The density is given in pounds per cubic foot. Since we want to find the specific heat capacity, which is the specific heat multiplied by the mass per unit volume, we need to convert the density to pounds per unit volume. 1 ft ³ = 12 in × 12 in × 12 in = 1728 in³ So,
Step 4:: Calculate specific heat capacity
C_p = 1.0 \frac{Btu}{lb°F} \times 0.036 \frac{lb}{in^3}
Substituting the values, we get:
Step 5:: Simplify the expression
C_p = 0.036 \frac{Btu}{in^3°F}
Final Answer
The specific heat capacity of water is 0.036 Btu/in³°F.
Need Help with Homework?
Stuck on a difficult problem? We've got you covered:
- Post your question or upload an image
- Get instant step-by-step solutions
- Learn from our AI and community of students