The heat required to heat a substance is given by the formula: Q = mcΔT, where Q is the heat energy, m is the mass of the substance, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature (50°C - 25°C = 25°C). Plugging in the values and rearranging for mass, we get: m = Q / (cΔT) = 2825 J / (4.18 J/g°C * 25°C) = 26.98 grams. So, the mass of water that can be heated from 25⁰C to 50⁰C by the addition of 2825 J of heat is approximately 27 grams.
When tin is heated in air, it can react with oxygen to form tin dioxide (SnO2). This compound has a higher mass than the original tin due to the addition of oxygen atoms from the air. As a result, the overall mass of tin increases when it undergoes this chemical reaction with oxygen in the air.
Only an insignificant amount, due to mass-energy equivalence - the added energy implies some added mass.
To find the mass of heated air, you need to know its volume, temperature, and pressure. Use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles of air, R is the ideal gas constant, and T is the temperature in Kelvin. By rearranging this equation and solving for mass (m = n * molar mass), you can calculate the mass of the heated air.
When a balloon is heated, the air inside the balloon expands and the molecules move faster, increasing their kinetic energy. This results in a decrease in mass density, as the molecules are more spread out. However, the total mass of the air inside the balloon remains the same.
When solids are heated, the atoms or molecules within them vibrate faster and with greater energy, causing them to spread out slightly. This increase in vibration leads to the expansion of the solid even though the mass remains the same.
To find the mass of water that can be heated from 25.0 °C to 50.0 °C with 2825 J, we can use the formula ( q = mc\Delta T ), where ( q ) is the heat energy (2825 J), ( m ) is the mass of water, ( c ) is the specific heat capacity of water (approximately 4.18 J/g°C), and ( \Delta T ) is the temperature change (50.0 °C - 25.0 °C = 25.0 °C). Rearranging the formula for mass gives ( m = \frac{q}{c\Delta T} = \frac{2825 , \text{J}}{4.18 , \text{J/g°C} \times 25.0 , \text{°C}} ). Calculating this yields a mass of approximately 27 grams of water.
None. Changing water from 25 degrees C to 5 degrees C requires heat to be REMOVED, not added!
no
use a thermetor
As a hydrated compound is heated, the water molecules contained within it will evaporate, leading to a decrease in mass. This process is known as dehydration.
Mudslides and landslides.
Mudslides and landslides.
It does not decrease in mass, only weight, the mass is still all there, but as when it has been heated smoke is given off from the element, and goes into the atmosphere. It may not sound like the mass is all there with the heated element, but if the smoke were conserved during the experiment the weight would be the same as before the experiment.
When tin is heated in air, it can react with oxygen to form tin dioxide (SnO2). This compound has a higher mass than the original tin due to the addition of oxygen atoms from the air. As a result, the overall mass of tin increases when it undergoes this chemical reaction with oxygen in the air.
Well, that might be thermal mass inside a dwelling, or water, as in hot water, heated by the sun.
Pure MgO will absorbe water from the air causing an error in the experiment if that extra mass is not accounted for. By adding water in the very beginning you eliminate this error.
Solid copper appears to gain mass when heated in air, because the copper reacts with oxygen in the air to form copper oxide. However, the actual mass of the copper does not increase; the mass of the solid increases by a value equal to the mass of oxygen removed from the air.