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How many btu must be removed from one pound of water at 200F for it to end as ice at 30F?
To calculate the heat energy required to cool water from 200°F to 32°F (its freezing point) and then to further cool it to 30°F and freeze it, we need to consider the specific heat capacities and latent heat of fusion of water. The specific heat capacity of water is 1 BTU/lb°F, and the latent heat of fusion of water is 144 BTU/lb. To cool water from 200°F to 32°F, we need to remove: (200°F - 32°F) * 1 BTU/lb°F = 168 BTU. To freeze the water at 32°F, we need to remove the latent heat of fusion: 144 BTU. Therefore, the total heat energy required to cool one pound of water from 200°F to ice at 30°F is 168 BTU + 144 BTU = 312 BTU.
How much heat is required to warm 25 g of water from 20deg cto 32deg c?
q=(275)(11)(1.00) q=3,025 cal remember that the formula to find heat is: q=m(DT)Cp and remember that the specific heat of water is: 1.00 cal/(gxC) you just replace values hope this helps:)
How many degrees does a liquid have to be in order to melt ice?
To melt ice, a liquid typically needs to be at least 0 degrees Celsius (32 degrees Fahrenheit). This is the freezing point of water, and any liquid at this temperature or warmer can facilitate the melting of ice. However, the efficiency of melting can vary based on the specific heat capacity and thermal conductivity of the liquid in question.
How many onces in a quart?
A quart is a unit of capacity. An ounce is a unit of mass. The two units are not directly compatible.
Calculate the specific heat of a substance if 44.0 g of the sample absorbs 67.2 J and the temperature change is 11 K?
The specific heat can be calculated using the formula: ( q = mc\Delta T ), where ( q ) is the heat absorbed, ( m ) is the mass, ( c ) is the specific heat, and ( \Delta T ) is the temperature change. Plugging in the values: ( 67.2 = (44.0) \times c \times 11 ), solving for ( c ) gives a specific heat of approximately 0.138 J/g∙K.
If a 4.0 g sample of glass was heated from 274 k to 314 and was foundd to absorb 32 j of energy as heat calculate the specific heat of this glass?
The specific heat capacity of glass can be calculated using the formula q = m * C * ∆T, where q is the heat absorbed (32 J), m is the mass (4.0 g), C is the specific heat capacity, and ∆T is the change in temperature (40 K). Rearranging the formula to solve for C gives C = q / (m * ∆T), which equals 0.2 J/gK for the specific heat capacity of this glass.
What is the oil capacity for 2001 international 4300 dt 466?
If I remember correctly, I believe oil capacity is 32 qts.
How much oil is needed for a 1997 Cadillac Northstar 32 valve engine?
The oil capacity of the 4.6L 32 valve Northstar in a 1997 Cadillac is 7.5 quarts.
What is the oil capacity in the primary for 1998 Harley Davidson fatboy?
The oil capacity for the primary chaincase of a 1998 Harley-Davidson Fat Boy is approximately 1 quart (32 ounces) when filled to the proper level. It's important to check the owner's manual or service manual for specific guidelines on filling and maintenance. Additionally, the total oil capacity for the engine is about 3 quarts (with a filter change). Always ensure to use the recommended oil type for optimal performance.
What is the oil capacity of a cummins n14 engine?
The oil capacity of a Cummins N14 engine typically ranges from 34 to 38 quarts (approximately 32 to 36 liters), depending on the specific model and configuration. It's important to consult the engine's manual for the exact specifications and requirements, as variations can occur based on factors like the oil filter and oil cooler. Regular maintenance and proper oil levels are crucial for optimal engine performance.
Calculated the energy to heat 32 grams of liqiud water from 47c to boiling?
To calculate the energy needed to heat the water, you can use the specific heat capacity of water, which is 4.184 J/g°C. Calculate the temperature change: boiling point of water (100°C) - initial temperature (47°C) = 53°C. Calculate the energy: Energy = mass (32 g) x specific heat capacity x temperature change. Substituting the values, we get Energy = 32 g x 4.184 J/g°C x 53°C = 7068.096 J or 7.07 kJ.
How many BTU's to raise water temp from 32 F to 212 F?
To raise the temperature of water from 32°F to 212°F, you would need 180 BTUs per pound of water. This calculation is based on the specific heat capacity of water, which is 1 BTU/lb°F.
How many kilocalories of heat are needed to raise the temperature of 460 grams of water from 13 C to 45 C?
The specific heat capacity of water is 1 calorie/gram °C. To convert this to kilocalories, we divide by 1000, so 1kcal/kg°C. The temperature change is 45 - 13 = 32°C. Therefore, the heat required is 460g * 32°C * 1kcal/kg°C = 14720 kcal.
What is the oil capacity of an 8V71 Detroit?
30 to 32 liters of CF2 40wt oil. Delo 100 40wt is best. Try to avoid multigrade 15w40.
How much oil goes in the primary of a 2003 fat boy?
The primary oil capacity for a 2003 Harley-Davidson Fat Boy is approximately 1 quart (32 ounces). It's important to use the recommended type of primary oil for optimal performance. Always check the owner's manual for specific guidelines and maintenance recommendations.
What is the fluid capacity in 1990 harley flt primary?
The fluid capacity for the primary chaincase on a 1990 Harley-Davidson FLT is approximately 32 ounces (or about 1 quart) of primary fluid. It is important to use the recommended type of fluid, typically a specific motorcycle primary oil, to ensure proper lubrication and function. Always check the owner's manual for specific details and recommendations.
How many btu must be removed from one pound of water at 200F for it to end as ice at 30F?
To calculate the heat energy required to cool water from 200°F to 32°F (its freezing point) and then to further cool it to 30°F and freeze it, we need to consider the specific heat capacities and latent heat of fusion of water. The specific heat capacity of water is 1 BTU/lb°F, and the latent heat of fusion of water is 144 BTU/lb. To cool water from 200°F to 32°F, we need to remove: (200°F - 32°F) * 1 BTU/lb°F = 168 BTU. To freeze the water at 32°F, we need to remove the latent heat of fusion: 144 BTU. Therefore, the total heat energy required to cool one pound of water from 200°F to ice at 30°F is 168 BTU + 144 BTU = 312 BTU.
