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A container has 2.4 liters of water at 20^C. What is the amount of heat required to boil the water? Write final answer to the nearest whole number and indicate the unit in kJ.
2.4l= 2400mldensity= 1ml/gMass= 1× 2400= 2400g= 2.4kgWater boils at 0°CBoiling point = 100°CInitial = 20°CSpecific heat capacity of water = 4200j/kg/°C= 4200×(100-20)×2.4= 806400j= 806.400kJ
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How much heat in calories and kilocalories, does it take to raise the temperature of 814g of water from 18.0°C to 100.0°C.
Heat needed =7,208.78cal
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Calculate the heat change of calories to condense 10.0 g of steam at 100 oC, and determine whether heat is absorbed or released
Enthalpy of vaporization for water is 2260 kJ/kg. Thus, the amount of RELEASED heat will be Q = r*m = 2260 kJ/kg*0.010 kg = 22.6 kJ or 5.4 kcal.
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If 4520 kJ of heat is needed to boil a sample of water, what is the mass of the sample of water
q = 4520 kJ = 4520000 J.Solution:For water at its normal boiling point of 100 ºC, the heat of vaporization (Hv) is 2260 J g-1. This means that to convert 1...
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how many calories were in the peanut if the following data are collected? mass of peanut burned= 0.25g volume of water heated = 100mL temperature change = 20'C specific heat of water = 1.00 cal/g'C
The density of water is 0.998 g/mL. Therefore the mass = 100 x 0.998 = 99.8 g.q = cm(T2 - T1) = 1.00cal/goC x 99.8g x 20oC = 1996 calAnswer:...
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Calculate the heat needed at 0∘C∘C to make each of the following changes of state.
Calories to melt 90 g of ice
80 calories needed to melt 1gm ICE at 0 degree.Hence for 90 gm ICE, it require 80x90 = 7200 calories So answer = 7200 calories.
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Calculate the heat needed at 0∘C∘to make each of the following changes of state.
Calories to melt 22 g of ice
QUESTION # 132550Calculate the heat needed at 0∘C∘to make each of the following changes of state.Calories to melt 22 g of iceANSWERSΔHf = q/mWhere;ΔHf = enthalpy of fusion of water(80cal/gram)q...
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What are the conversation factors for the equality 1 meter = 100 centimeters?
a. 1 m /100 cm and 100 cm/1m
b.1 m/100 cm
c. 1 cm/100 m and 100 m/1 cm
d. 1 m/1000 cm and 1000 cm/1 m
Solutions: Meters (m) and centimeters (cm) can be related by the prefix modifier equality 100 cm = 1 m.So therefore, this equality given above can be represented as two conversion...
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Calculate the heat change in calories for condensation of 13.0 g steam at 100 C.
Please express answer as a positive value using three significant figures and include the appropriate units.
Specific latent heat of vaporization of water at 100°C = 2260 J/gHeat change = (13.0 g) × (2260 J/g) = 29380 J = 7022 cal;Answer: 7022 calories
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You need to heat 7.59 g of water from 5.2 ºC to 368 K. How much heat is needed? Give answer in Joules and calories. _____J _____cal
5.2°C = (273 + 5.2)k = 278.2kChange in temperature = 368k -278.2= 89.8kq (heat energy) = MCp∆TWhen Cp of water =1046J q = 7.59 g × 1046J × 89.8k =...
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