What is the final temperature of the mixture when 1 kg of water at 10° C is mixed with 4.4 kg of water at 30°C?
The question is wrongly phrased. Ice can not be at 10°C. So it should be worded “ What is the final temperature of the mixture when 1 kg of ice is mixed with 4.4 kg of water at 30°C?”OrWhat is the temperature of the mixture when 1 kg of water at 10° C is mixed with 4.4 kg of of water at 30°C?I shall solve both one by oneCase 1: 1 kg of water at 10° CLet the final temperature be T° C , where T°C < 30°C.Heat give out by 4.4 kg of water at 30° C to come down to T°C = 4400 g × 1 calorie/g/°C × (30- T)° C = (4400× 30 - 4400 T) calorieHeat gained to raise temperature of 1kg of water at 10° to T° = 1000 g × 1 calorie/g/°C × (T -10)° C = 1000 T - 10,000) calorieHeat gained = Heat lost1000 T - 10,000 = 1,32,000 - 4400 T; ==> 54 00T = 1.42,000; T = 1,42,000/5400 = 26.3° C.In case it is 1 kg of water at 10° C mixed with 4.4 kg of water at 30° C, the final temperature of the mixture would be T = 26.3° C.Case 2: 1 kg of ice.Ice is essentially at its melting point/freezing point ie at 0° C.Let the temperature of mixture = T° CHeat required to melt 1 kg (=1000 g of ice) at 0° C to water at 0° C = 1000 g × 80 calorie/g = 80,000 calorieHeat require to raise temperature of 1000 g of water at 0° C to water at T° C = 1000g × 1 calorie/g/°C× (T -0)° C = 1000 T calorieHeat gained = 80,000 + 1000 THeat lost by 4.4 kg of water at 30° C to cool to T° C = 4400 g × 1 calorie/g /° C × ( 30 - T)° C = 1,32,000 - 4400 THeat gained = Heat lost80,000 + 1,000 T = 1,32,000 -4400 T5400 T = 52,000/5400; ==> T = 9.63°C.In case it is ice, the temperature of the mixture is T = 9.63° C.Added: around 2 pmIn the original post the question was , “ What is the final temperature of the mixture when 1 kg of ice at 0° C is mixed with 4.4 kg of water at 30° C.”We know ice cannot be at 10° C. If the temperature is 10° C then it is not ice it is already water at 10°C. If it is ice then we need not say the temperature. Temperature of ice is always taken as 0° C, if not specified. But it cannot be taken as 10° C. SSo I edited the question. And answered both possible case.
If more ice is added to an ice-water mixture at equilibrium...?
If more ice is added to an ice-water mixture at equilibrium... a) the vapor pressure of the water will decrease b) the temperature will increase somewhat c) the temperature will decrease somewhat d) the vapor pressure of the water will rise e) the vapor pressure of the water will remain constant
How to do you determine the final temperature of aluminum added to a mixture of ice and water?
Lets just do the problem and then you can do some comparing. The first issue is a "mixture" of ice at 0C and water and 95C. Such a mixture will rapidly come to thermal equilibrium as the ice melts and cools the water. In the equation below we use the term "ice water" to represent the water formed from the melting ice in addition to the water that was already present in the mixture. q lost by water as it cools to Tf = q gained by ice as it melts + "ice water" as it warms to Tf -mcΔT(hot water) = mHv + mcΔT(ice water) -2000.0g x 4.18J/gC x (Tf - 95.0C) = 45.0g x 334 J/g + 45.0g x 4.18J/gC x (Tf - 0) (algebra occurs here) Tf = 91.15C ..... this is simply the starting temperature for what was the mixture of ice and water. Next we mix in the aluminum.... q gained by Al as it rises to Tf = -q lost by hot water as it cools. mcΔT(Al) = -mcΔT(hot water) 45.8g x 0.900 J/gC x (Tf - (-5.2)) = -2045g x 4.18J/gC x (Tf - 91.15C) (algebra occurs here) Tf = 90.7C ...... consistent to one decimal point
A 120 g insulated aluminum cup at 20°C is filled with 240 g of water at 100°C.?
We need to find the final temperature of the system. Assuming no losses, the cup and water will end up at the final temperature T₂ Specific Heat from link C_aluminum = 0.215 C_water = 1.00 Q_cup + Q_water = 0 M_cup C_cup ΔT_cup + M_water C_water ΔT_water = 0 120g * 0.215 * ΔT_cup + 240g * 1 * ΔT_water = 0 25.8 ΔT_cup + 240 ΔT_water = 0 25.8 (T₂ - T₁_cup) + 240 (T₂ - T₁_water) = 0 25.8 T₂ - 25.8 T₁_cup + 240 T₂ - 240 T₁_water = 0 265.8 T₂ = 25.8 T₁_cup + 240 T₁_water T₂ = (25.8 T₁_cup + 240 T₁_water)/265.8 = (25.8 * 20°C + 240 * 100°C)/265.8 = 92.2°C 20°C = 293°K 92.2°C = 365.2°K 100°C = 373°K ΔS = ΔS_cup + ΔS_water = Q_cup/T_avg + Q_water/T_avg = (MCΔT)_cup/T_avg + (MCΔT)_water/T_avg = (120g * 0.215 * (365.2°K - 293°K))/((365.2°K + 293°K)/2) + (240g * 1 * (365.2°K - 373°K))/((365.2°K + 373°K)/2) = 1862.76/392.1 + -187.2/369.1 = 0.588
2 kg of ice at 0°c is mixed with 8 kg of water at 20°c. What is the final temperature?
Final temp. Will be water at 0 degree celcius
If 10 g of ice at - 10° C is added to 50 g of water at 15° C, what is the temperature of the mixture?
You need to understand 3 concepts to solve this answer.Firstly, You have to use the principle of calorimetry (measuring heat content of a system) to get the answer for this question.The principle goes like this:heat gained by a system(ice) = heat given by a system (water).That means the heat absorbed by ice will be equal to the heat given by the water.Secondly,now since you know the principle, next step is to know the formula of heat given or taken, that isQ= m*c*(t2-t1)where,m=mass of the substancec=heat capacity of substance (ice=2.1kJ/kgK, water=4.2kJ/kgK)t2-t1=change in temperature of the substanceNow the last but not the least concept, you must know what is latent heat, you see while there is a change of state that is when solid converts to liquid or liquid to water there is no change in temperature with the absorption of heat and that value for ice to water is 336kJ/kg.So this is the process is which is gonna happen, as the ice comes in contact with water, its temperature will rise upto 0 deg celsius. At that temperature which is the melting point of ice, it converts into water without any change in temperature. After complete conversion it again absorbs heat to reach a state where the converted ice and the water are at same temperatures (thermal equillibrium).So your final formula would be:Heat given by water = heat absorbed by ice + latent heat + heat absorbed by water (melted ice)mw*cw*(tw-T) = mice*cice*(tice-0) +336 +mmw*cmw*(T-0)The only unknown here is T =mixture temperature.Hope you got the science. :)
If 100g of ice at 0 degree celcius is added into 100g of water at 80 degree celcius, what is the final temperature of the mixture?
Let t is final temperature of the mixture then heat taken by100gm of ice to melt=mL=100*80=8000 cal heat taken by water of 0 degree to raise it's temperature to t degree =mst=100*1*t=100t thereforetotal heat required by ice=8000+100theat given by 100gm of water at 80 degrees=100*1*(80-t) =100(80-t)heat taken=heat given8000 + 100t=100(80-t)8000+100t=8000–100t200t=0t=0 dergee Celsius. AnswerExplaination :- calories needed for ice to meltto become water at 0 degree =8000 caloriesMaximum heat which 100 gm. ofwater can give=100*1*(80-0)=8000 degreesso maximum heat of 8000 cal. which water can givewill be utilized only to melt the iceand in this process both water due to ice and the given water both will come at 0 degree Celsius.Hence temperature of mixture will become zero degree Celsius.
800g of ice at -15°c is mixed with 100g of water at 40°c. What is the final temperature when equilibrium is attended?
I am extremely sorry for my handwriting. If there's something that you can't understand, just leave a comment and I'll explain. Again I am really sorry for the handwriting.Essentially the concept here is the fact that after the ice melts, the mixture will have a temperature T that will be lower than the initial water temperature (40° C). And according to the laws of thermodynamics, the heat gained by the ice will be equal to the heat lost by the water. Now the ice will gain heat in three steps viz. By heating up from -15 C to 0 C, melting and gaining it's latent heat and then heating up from 0 C to mixture temperature T° C. And the water will lose heat in a single step viz. By cooling down from 40 C to T° C.I haven't provided an exact answer, just the basic equation that only needs a little algebraic manipulation to obtain the answer. I hope this helps.Thanks for the A2A anon. ;)