How will the internal energy of the transformation of 500 g of ice taken at a temperature of 0 ° С change into water having a temperature of 20 ° С? Disregard the energy losses for heating the ambient air. 1) decrease by 42 kJ 2) increase by 42 kJ 3) decrease by 207 kJ 4) increase by 207 kJ
m = 500 g = 0.5 kg.
t1 = 0 ° C.
t2 = 20 ° C.
q = 3.4 * 10 ^ 5 J / kg.
C = 4200 J / kg * ° С.
The amount of heat Q that is imparted to the ice to transform it into water is used to increase its internal energy ΔU: ΔU = Q.
The ice must first be melted, and then the resulting water must be heated.
The amount of heat required for this is found by the formula: Q = q * m + C * m * (t2 – t1) = m * (q + C * (t2 – t1)).
Q = 0.5 kg * (3.4 * 10 ^ 5 J / kg + 4200 J / kg * ° C * (20 ° C – 0 ° C)) = 212000 J = 212 kJ.
Answer: the internal energy will increase by ΔU = 212 kJ.
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