A 4.2-ke block of ice originally at 263 K is placed in thermal contact with a 13

Question

A 4.2-ke block of ice originally at 263 K is placed in thermal contact with a 13.7-kg block of silver (CAg = 233 j/kg-K) which is initially at 1086 K. The H2O - silver system is insulated so no heat flows into or out of it. 1) At what temperature will the system achieve equilibrium? You currently have O submissions [or this question. Only 10 submission are allowed. You can make 10 more submissions (or this question. 2) What will be the phase of the H20 at equilibrium? Solid (ice) Liquid (water) Gas (vapor) Submit You currently have O submissions (or this question. Only 10 submission are allowed. You can make 10 more submissions (or this question.

Explanation / Answer

Let asssume that the ice turned into water

Hence heat required = Q = msicedt1+mL+mswaterdt2

letequilibrium temperature be T

Q = 4.2(2100*(273-263) + 4190*(T-273)) = 88200+17598T-4804254 = 17598T-4716054

heat given off by silver block = Q = msagdt = 13.7*233*(1086-T) = 3466620.6-3192.1T

at equlibrium heat taken by ice = heat given by silver

17598T-4716054 = 3466620.6 - 3192.1T

20790.1T = 8182674.6

T = 393.6 K (Here T is an acceptable value. If in such problems, T is unaccepatable value like more than 1086 or negative then we have consider that the final phase is steam and proceed again)

2. since T = 393.6K is acceptable value, at equilbruim, phase of H2O is liquid

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