A well-insulated rigid vessel contains 3 kg of saturated liquid water at 40oC. The vessel also contains an electrical resistor that draws 10 amperes when 50 volts are applied. Determine the final temperature in the vessel after the resistor has been operating for 30 minutes. ( 119oC)

The final temperature is 111.66°C

Explanation:

The given conditions :-

i) Well insulated means no heat loss.

ii) Rigid vessels means volume remains same.

iii) Initial temperature ( T₁ ) = 40°C. = 273 + 40 = 313 K

iv ) Mass of water in vessel = 3 kg.

v) current drawn by resistor ( i ) = 10 ampere.

vi) Voltage applied ( V ) = 50 volts.

vii) The time for which resistor operating ( t ) = 30 minute = 30 * 60 = 1800 seconds.

Now we have to calculate heat developed by resistor in vessel.

Q = V * i * t  = 50 * 10 * 1800 = 900,000 J = 900 KJ.

Since it is a rigid container so the work done is zero.

Q = du    ( du - change in internal energy)

Q = m * C * dT      ( C = 4.186 KJ/KgK )

Q = 3 * 4.186 * (T₂ - T₁ )

900 = 12.558 * ( T₂ - 313 )

T₂ - 313 = 71.6674

T₂ = 384.6674 K

T = 384.6674 - 273 = 111.66°C

So the final temperature is 111.66°C.

answer:

the answer is d

explanation:

answer:

explanation:

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