Question

Favor ayudarme lo mas pronto posible con la pregunta 3 solo literal b. Gracias.

Answer 1

Q = m Ce (T- t2)

Ce agua = 1 cal/gºC

Ce plata = 0.056 cal/g 1ºC

m1Ce (T- t2) = m2Ce (T- t2)

800 g x 1 cal/g ºC ( x – 15 ºC) = 300 g x 0.056 cal/g ºC ( 80ºC –  X)

800 Cal /ºC – 12000 g/cal = 1344 cal – 16.8 cal/ºC

800 (+ 16.8 ) Cal/ºC = 1344 cal. + 12000 cal.

816.8 cal/ºC = 13344 Cal.

ºC = 13344 cal / 816.8 cal

ºC = 16.34