500*20+500*80+500*250+500*100+500*537+500*150*0.5=1000+4000+125000+50000+268500+37500=486000 Calories=468 K.cal.
155*25+155*80+155*100+155*536+155*150 calories=138105 calories=138.105 Kilo calories.
52.5 kcal
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.
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
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
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
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)
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).
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
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.
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.