How much heat is required to warm 1.50L of water from 20.0∘C to 100.0 ∘C? (Assume a density of 1.0g/mL for the?
A calorie is commonly defined as the amount of heat required to raise the temperature of one gram of water 1 degree C. You have 1.5 liters or 1500 grams of water, and you want to raise the temperature 100 - 20 = 80 degrees C. 1500 x 80 = 120000 calories If you need a different unit for heat: 1 kcal = 4186.8 J = 426.9 kp.m = 1.163 10-3 kWh = 3.088 ft.lbf = 3.9683 Btu = 1000 cal
How much heat is required to warm 1.50 L of water from 22.0 ∘C to 100.0 ∘C? (Assume a density of 1.0g/mL for the water.) in joules?
(1500 mL) x (1.0 g/mL) x (4.184 J/g·°C) x (100.0 - 22.0) °C = 4.9 x 10^5 J
How much heat is required to warm 1.50L of water from 28.0∘C to 100.0 ∘C? (Assume a density of 1.0g/mL for the?
How much heat is required to warm 1.50L of water from 28.0∘C to 100.0 ∘C? (Assume a density of 1.0g/mL for the water.) I tried to do this problem myself several times but I shouldn't couldn't get it. I followed the formula q=m*c*delta T so I plugged in 1500*-285.8*72 But the answer I got was wrong. The answer is 4.5×10^5 J How did they get that answer?
How much heat is required to warm 1.80L of water from 30.0∘C to 100.0∘C? (Assume a density of 1.0g/mL for the water.)?
1.8L is 1,800 ml. It takes one calorie to raise on ml of water by 1 degree C. You need a 70 degree rise. That's 1,800 times 70 = 126,000 calories, which is 126 kilocalories. This is 527,184 joules.
How much heat is required to warm 1.40 L of water from 30.0 C to 100.0 C? density 1g/mL?
Amount of heat= 410,032 J. I need to put it on the top so you'll read my steps. :) The formula that you use for this problem is Amount of heat (J)=Mass(g)*Heat Capacity (J/g*celcius)*Temperature Change (Celsius). Apply this formula to the problem. You are trying to find the amount of heat required to warm the water. 1. Convert the 1.40 L of water to mL. That would be 1400 mL of water. 2. density is 1 g/mL. You are trying to find how many g of water. 1=g/1400 mL. Solve to find g. That would be 1400 g. Another thing is, 1 mL of water will *always* be equal to 1 gram of water. Only water works in this case ^-^ 3. You should know that the heat capacity of water is 4.184 J/g*Celsius. 4. Apply formula. The shortened formula that you can write quickly is q=m*c*delta T for tests and stuff. I'll solve it for you: Amount of heat=1400 g * 4.184 * (100-30 C) Amount of heat= 410,032 J That's a freaking lot of energy o_o But i've checked it :)
How much heat is required to warm 1.70 L of water from 30.0^C to 100.0 ^C}?
I'm assuming you mean how much energy is required to raise water from 30c to 100c. That is an area of thermaldynamics known as specific heat. Q(heat) = c(specific heat) * m(mass) * dT (change in temperature) The c (specific heat) value for water is 4.186 joule/gram. The m (mass) is 1.7kg or 1700 grams. The dT (change in temperature) is 70 degrees (c). Q(heat) = 4.186 J/gc * 1700 g * 70 c Q(heat) = 498134 J The one catch is that this only works for liquid water and 100 degrees is the edge of converting it to gas.
How much heat is required to warm 1.40 L of water from 30.0 C to 100.0 C? density 1g/mL?
You find the change in temperature by subtracting you bigger temperature by the smalled, which would be 70 degree celcius. After that convert your 1.40L to grams using this formula: 1.40 * 1000ml/1L *1L/1g...which will give you 1400g. using the heat formula of q= m(mass) * Cs (specific heat capacity) * change in temperature. The specific heat capacity of water is 4.18 J/g*C...you equation should look like this: q=1400*4.18*70...u final answer will be 409640 or 4.1×10^5
How much heat is required to warm 1.60 L of water from 30.0 C to 100.0 C? Assume a density of 1.0 g/mL H2O?
One calorie = the amount of heat needed to raise the temperature of 1.0 milliliter of water by one Celcius degree. So: 1.60 liters = 1600 milliliters 100°C - 30°C = 70 Celcius degrees 70 * 1600 = 112000 calories and you can convert that into whichever equivalent units are desired by the question
How much heat is required to warm 1.50L of water from 30.0∘C to 100.0 ∘C?
Heat required = [mass] * [specific heat] * [change in temperature] mass = volume * density = 1500ml * 1.0g/ml = 1.50kg change in temperature = (100 - 30) = 70 degrees You haven't given us your value for specific heat. I'll use s = 4.187 kJ/kg.∘C Heat required = 1.50kg * 4.187 kJ/kg.∘C * 70∘C = 439.64 J = 440 J
How much heat is required to warm 1.70 L of water from 22.0 C to 100.0 C?
heat = (change in temp) * mass * specific heat Change in temp = 100 - 22 = 78 degrees C Mass = 1.70 L * 1000 mL / 1 L * 1 g / 1 mL = 1700 grams. Specific heat of water is 4.186 Joules per gram degrees C So heat = 78 C * 1700 grams * 4.186 Joules per gram degrees C The degrees C and grams cancel out leaving Joules as your unit. 555063.6 Joules, but it may be more common to put it in kiloJoules. Divide that number by 1000 and you get 555.06 kJ as your answer.