What would the expected temperature change be (in fahrenheit) if a 0.5 gram sample of water released 50.1 j of heat energy? the specific heat of liquid water is 4.184 j/g-c

75.11°F

Explanation:

Given data:

Change in temperature = ?

Mass of water = 0.5 g

Heat released = 50.1 J

Specific heat of water = 4.184 j/g. °C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Q = m.c. ΔT

50.1 J = 0.5 g . 4.184 j/g. °C. ΔT

50.1 J = 2.092 j/ °C .ΔT

50.1 J / 2.092 j/ °C  = ΔT

23.95 °C =ΔT

°C to °F

(23.95°C ×9/5 )+32 = 75.11°F

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.