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

The vapor pressure at 28 Co is of 28.3 mm Hg. Based on the Dalton’s law regarding partial pressure of gas mixture the total pressure is defined as follows: P...

Heat needed =7,208.78cal

Q = cmΔT ΔT = 19.2 – 14.4 = 4.8˚C = 4.8 K C (water) = 4.218 kJ/(kg·K) Q = 4.218 · 0.075 · 4.8 = 1.52 kJ

Solution:Assume that there are no phase transitions (solid → liquid and liquid → gas) and mercury (Hg) remains liquid in the whole temperature range (from -40°C to 356.6°C).q = m...

Solution:q = m × S × ΔTq = m × S × (Tf - Ti),where:q = amount of heat energy gained or lost by substance (J)m = mass of sample (g)S =...

To heat 1 g on 1 degree C we need 1 cal. So to heat 37 g on 78 degrees C we need: C=37*78=2886 cal. 1 cal.=4.184 J So, 2886...

q = m x C xΔT q = 250g x 4.18J/goC x (98 – 22)oC q = 79.4 kJ

∆H = mC∆Tm = 56.1 g;C = 4.18 J/(g×°C);∆T = 22.5 - 14 = 8.5 °C;∆H = 56.1 g × 4.18 J/(g×°C) × 8.5 °C = 1993.2 J;Answer: 1.99 kJ

80 calories needed to melt 1gm ICE at 0 degree.Hence for 90 gm ICE, it require 80x90 = 7200 calories So answer = 7200 calories.

Heat of water vaporization 2260 kJ/kg. 1 kJ = 0.239 kcal, 2260 kJ/kg = 540.14 kcalQ = 540.14*0.155 = 83.7217 kcal = 83721.7 cal.