A 30.0-g sample of water at 280. K is mixed with 50.0-g of water at 330. K. Caculate the final temperature of the mixture assuming no heat is lost to the surroundings.
2007-10-21 05:04:50 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
The specific heat of water is 4.18J/C*g
2007-10-21 05:09:26 · update #1
No specific heat of water is needed. Please use ALGEBRA!
Let the final temperature of the mixture be T.
30.0(T - 280) = 50.0(330 - T)
80T = 24900
T = 311.25 (K)
2007-10-22 08:40:03 · answer #1 · answered by Hahaha 7 · 0⤊ 0⤋
i'm going to call ethylene glycol "EG." warmth required = (mass EG)(sp. ht. EG)(delta T) = (sixty two.0 g)(2.forty two J/g C)(forty.5 C - 13.3 C) = 4080 J ================================== The equation says that a million mole of AgNO3 reacts with a million mole of HCl to offer -sixty 8 kJ of warmth. fifty six.7 g AgNO3 x (a million mole AgNO3 / 169.9 g AgNO3) = 0.334 moles AgNO3 0.334 moles AgNO3 x (-sixty 8 kJ warmth / a million mole AgNO3) = -23 kJ warmth
2016-11-09 02:37:26 · answer #2 · answered by fones 4 · 0⤊ 0⤋