For 1 gram of water, you need 4.184 Joule to raise the T from 15 degrees Celsius to 16 degrees Celsius. The values are slightly different for other temperatures. So an estimation of the required energy would be 10 * 4.2 * 80. The 10 comes from 10 gram, the 4.2 from the energy required to heat 1 gram of water 1 degrees Celsius. The 80 comes from the starting temperature to the final temperature. The estimated answer is then 3360 J. For 1 gram of water, you need 4.184 Joule to raise the T from 15 degrees Celsius to 16 degrees Celsius. The values are slightly different for other temperatures. So an estimation of the required energy would be 10 * 4.2 * 80. The 10 comes from 10 gram, the 4.2 from the energy required to heat 1 gram of water 1 degrees Celsius. The 80 comes from the starting temperature to the final temperature. The estimated answer is then 3360 J.
10 g of ice has to melt at 10 degrees C and the calories required would be 10 times the latent heat of fusion of ice which is approximately 80 cals/gram. Then we have to rise the temperature of 10 g of water from 10 degrees C to 20 degrees C.
This would be 10 x 10 x specific heat of water which is 1 cal per gram
The total energy in calories required would be
(10 x 80) + (10 x 1 x 10) = 800 + 100 = 900cals
10*10 calories to raise the temp of ice from -10 deg C to 0 deg C
10*80 calories to melt the ice to water
10*20 calorie to raise the temp of water from 0 deg C to 20 deg C.
= 100 + 800 + 200 = 1100 cals.
Change ice to water- 10g ice to water
Equation: Q= m x specific latent heat of ice
Q = 10g * 340
Q = 3400J
The specific heat of ice is 2,05 J/g.K
The specific heat of water is4,186 J/g.K
So the total necessary heat is:
Q = (2,05 x 10 x 1o) + (4,186 x 10 x 20) = 1 042,2 joules
The amount of heat needed to raise temperature of a substance is measure by the equation Q=mc@. The difference of temperature is 30 degrees Celsius. Therefor Q=0.3J.
The specific heat capacity of ice is 2,05 J/g.K.
The specific heat capacity of water is 4,186 J/g.K.
Necessary heat is: (2,05 x 10 x 10) + (4,186 x 20 x 10) = 1042,2 J
100 calories
6300 jules
You need to add all of the following:* The heat required to heat ice from -5 to 0 degrees. Multiply the mass times the temperature difference times the specific heat of ice. * The heat required to melt ice. Multiply the mass by the heat of fusion. * The heat required to raiste the temperature of water from 0 to 20 degrees. Multiply the mass times the temperature difference times the specific heat of water.
Water has a high specific heat capacity (relative to metals and other conductors), making it a poor conductor of heat (takes too much energy to change the temperature).
Heat is used in three stages 1. To rise the temperature of ice from -7 to 0 deg celsius 2. To change ice into water - melting 3. To rise temperature of water from 0 to 20 deg Celsius Hence Heat = 0.380*S*7 + 0.380*L+0.380*s*20 S - specific heat capacity of ice s- specific heat capacity of wate L= Laten heat of fusion of ice. Please get the data from data book, plug and find the heat needed
Heat of vaporization or enthalpy of vaporization. It is the additional energy, per unit mass, required after vaporization temperature (boiling point) is reached, to accomplish the change in state, from liquid to gas.
Because Water's latent heat of fusion is much less than its latent heat of vaporization. In English: It takes less energy to change a gram of ice at 0°C into a gram of water at the same temperature than it takes to change a gram of water at 100°C into a gram of steam at the same temperature.
To change the temperature of water from 27ºC to 32ºC will depend on the mass of water that is present. Obviously, the more water, the more heat it will take. This can be calculated as follows:q = heat = mC∆T where m is the mass of water; C is sp. heat = 4.184 J/g/deg and ∆T is 5ºC (change in temp).
The necessary heat is 9,22 joules.
The idea here is to: * Look up the specific heat of water. * Multiply the mass, times the temperature difference, times the specific heat of water. You may need to do some unit conversions first; specifically, if the specific heat is given per kilogram, you can convert the grams to kilograms.
You need to add all of the following:* The heat required to heat ice from -5 to 0 degrees. Multiply the mass times the temperature difference times the specific heat of ice. * The heat required to melt ice. Multiply the mass by the heat of fusion. * The heat required to raiste the temperature of water from 0 to 20 degrees. Multiply the mass times the temperature difference times the specific heat of water.
Heat required to have such a change of state is called latent heat. If L J/kg is the latent heat per kg of water then for M kg of water we need M* L joule of heat energy
water has a high heat of vapourization.it absorbs much heat as it changes from liquid to gas.it has the capacity of absorbing heat with minimum of change in its own temperature
Water has a high specific heat capacity (relative to metals and other conductors), making it a poor conductor of heat (takes too much energy to change the temperature).
The specific heat capacity of water does not change much within-phase (ie, as a solid it has one specific heat capacity, as a liquid/gas it has another)
because the specific heat of water is very high
Heat is used in three stages 1. To rise the temperature of ice from -7 to 0 deg celsius 2. To change ice into water - melting 3. To rise temperature of water from 0 to 20 deg Celsius Hence Heat = 0.380*S*7 + 0.380*L+0.380*s*20 S - specific heat capacity of ice s- specific heat capacity of wate L= Laten heat of fusion of ice. Please get the data from data book, plug and find the heat needed
The equation is q = mC∆T where q is the heat; m is the mass of water; C is the specific heat of water (1 cal/g/deg); and ∆T is the change in temperature.
Heat of vaporization or enthalpy of vaporization. It is the additional energy, per unit mass, required after vaporization temperature (boiling point) is reached, to accomplish the change in state, from liquid to gas.