I must assume that the ice block is at an initial temperature of 0o C. I also need to work in joules,. so I will convert the calories to joules.
1600 cal. (4.184 joule/1 cal.)
= 6694.4 joules
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6694.4 J = (20 grams H2O)(4.180 J/gC)(Tf - 0o C)
6694.4/83.6 = Tf
= 80o Celsius
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The final temperature would be approximately 54.2 degrees Celsius. This can be calculated using the principle of conservation of energy, where the heat lost by the hot water is equal to the heat gained by the cold water.
To find the final temperature, you can use the formula: heat gained = mass * specific heat capacity * temperature change. First, calculate the heat gained: 1200 calories = 40g * 1 cal/g°C * (final temperature - 20°C). Rearrange the formula to solve for the final temperature: final temperature = 1200 calories / (40g * 1 cal/g°C) + 20°C. Solving this gives the final temperature of 50°C.
The amount of water in fog over a city block can vary widely depending on factors like temperature, humidity, and fog density. However, on average, fog can contain around 0.05 to 0.5 grams of water per cubic meter. This means that over a city block, which can cover an area of 10,000 to 20,000 square meters, there could be between 500 to 10,000 grams (or 0.5 to 10 kilograms) of water in the fog.
Grams are mass. There's no calculation involved.
To calculate the grams of oxygen in 34.7 ml, we need to know the density of oxygen at the given conditions. The density of oxygen at room temperature and pressure is 1.429 g/ml. Multiply the volume by the density: 34.7 ml * 1.429 g/ml = 49.64 grams of oxygen.
The final temperature would be approximately 54.2 degrees Celsius. This can be calculated using the principle of conservation of energy, where the heat lost by the hot water is equal to the heat gained by the cold water.
The volume depends on the temperature and pressure - neither of which are given.
Do some converting first. 688 calories (4.184 Joules/1 calorie) = 2878.592 Joules 25 ml of water = 25 grams q(Jolules) = mass * specific heat * (Temp. final - Temp. initial) 2878.592 Joules = 25 grams Water * 4.180 J/gC * (Temp Final - 80C ) 2878.592 Joiles = 104.5( Temp. Final) - 8360 11238.592 =104.5(Temp. Final) 107.55 Celsius Final Temperature ( call it 108 C )
The final temperature can be calculated using the principle of conservation of energy. The formula to use is: ( m_1c_1(T_f-T_1) = -m_2c_2(T_f-T_2) ), where (m_1c_1) is the mass and specific heat capacity of water at 70ºC, (T_f) is the final temperature, (m_2c_2) is the mass and specific heat capacity of water at 10ºC, and (T_1) and (T_2) are the initial temperatures. Substituting the values, we find the final temperature to be around 24ºC.
It is not possible to answer the question in which the volume of the block is given as Cm Cm and Cm. Some numbers might have been useful!
As the volume of a given gas sample is dependent on its temperature and pressure; to find a volume of any gas which does exist, the temperature and the pressure of the system/vessel should be given directly or could be calculated.
To find the final temperature, we can use the formula: q = m x c x ΔT, where q is the heat energy, m is the mass of water, c is the specific heat capacity of water, and ΔT is the temperature change. By rearranging the formula and substituting the values, we can find the final temperature to be approximately 39.8°C.
To find the equilibrium temperature, we can apply the principle of conservation of energy using the formula: mcΔT = -mcΔT, where m is the mass, c is the specific heat, and ΔT is the change in temperature. Setting the two sides equal to each other and solving for the equilibrium temperature, we can find that the final temperature will be around 38.7°C.
About 232 grams
well you have to think you would weigh this using grams so the density of a foam block is "Grams per cubic centimeters"
The measure of how well a solute can dissolve at a given temperature is known as solubility. It is typically expressed as the maximum amount of solute that can dissolve in a specific amount of solvent under particular conditions, usually in grams of solute per 100 grams of solvent.
875J/12.6g*4.18Jg-1°C+22.9°C=39.5°C