2,092 J is the quantity of heat energy released when 50.0 grams the water is cooled.
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Mass of water = m = 50.0 g
Initial temperature of the water =
Final temperature of the coffee =
Specific warm of water = c = 4.184 J/g°C
Heat required to cooled from 20°C come 10°C= Q
Q = -2,092 J
Negative sign means that power is lost.
2,092 J is the lot of heat energy released once 50.0 grams the water is cooled.
The amount of warmth released as soon as 50 g of water cooled native 20°C come 10°C will certainly be equal to - 2093 J.
Mass the water = 50 g
Initial temperature= T1 = 20°C
Final temperature= T2 = 10°C
Specific warm of water= c = 4.186 J/g. °C
Amount of warmth released = Q= ?
Q = m. C. ΔT
ΔT = T2 - T1
ΔT = 10°C - 20°C
ΔT = -10°C
Now we will put the values in formula.
Q = m. C. ΔT
Q = 50 g . 4.186 J/g. °C . -10°C
Q = - 2093 J
The amount of warmth released as soon as 50 g of water cooled indigenous 20°C to 10°C will be equal to - 2093 J.
Q = mc Δtq= heat, m = fixed (in grams), c=specific warm (4.18 J/°C * g or J/K * g for water), Δt = readjust in temperature (in kelvin)For this equation:q = ?, m = 50.0 g, c = 4.18 J/°C *g = 4.18 J/K *g, early t = 20.0°C, last t = 10°CTo convert from °C to K, include 273 to °C0°C = 273 KInitial t = 20.0°C + 273 = 293 KFinal t = 10.0°C + 273 = 283 KΔT = last temperature - initial temperature = 283-293 = -10 KPlug in what you know:q = (50.0 g) x (4.18 J/K * g) x (-10 K) = -2090Check sig figs and cancel out units: -2090 J (you need 3 sig figs from the initial numbers given and also the units space Joules because grams and Kelvin cancel out)Check to make sure answer makes sense:Since heat power is released, the answer have to be negative. The price is an unfavorable and therefore, renders sense
q = mC∆T
q = (50 g)(4.184 J/g/deg)(10 deg)
q = 2092 J = 2.09x10^3 J
That"s how it"s done.
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