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To increase the temperature of 50 grams of water by 1 degree C requires how many calories?
Wiki User
∙ 11y ago
Q=mc(delta)T .In this problem T = the temperature -> 50 ºC ,m= 500 grams, c= specific heat capacity, for water which is 1 cal/gºC. Q= 500g x 1 cal/gºC x 50 ºC = 25,000 cal.
Wiki User
∙ 10y ago
Wiki User
∙ 13y agoIt should actually take 5000 calories.
Here is how:
Q = c x m x (Tf-Ti),
where c = specific heat, m = mass, Tf = final temp, and Ti = initial temp
or
(Heat) = (Specific Heat) times (Mass) times (Change in Temperature)
Thus,
Heat = 1cal/g *C x 100 g x 50*C = 5000 cal
Wiki User
∙ 14y agoQ=mass x (delta)T x specific heat capacity
50.0g x (83.0c-4.0c)x4.18j/gc
= 17000 J of heat with the appropriate significant figures
Wiki User
∙ 10y agoq(joules) = mass * specific heat * change in temperature
q = (20 grams H2O)(4.180 J/gC)(50o C - 25o C)
= 2090 joules of heat energy
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Wiki User
∙ 13y ago4.184 x 50 x 100 = 20920 Joules or 20.92 kJ
Wiki User
∙ 8y agoThe specific heat of water is 4.179 Joules per (gram-degree-Celsius).
So, the specific heat of 50 grams of water would be: 4.179 J/g-degC * 50 g = 208.95 J/degC
Wiki User
∙ 8y agoThe specific heat of water is 4181 J/kg.oC
Wiki User
∙ 11y agoA little over 1 calorie.
Wiki User
∙ 11y ago50 x 2 x 4180 = 418,000
Wiki User
∙ 15y agoAdding 41.84J
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