Question

Copper is sometimes coated on the bottom surface of cooking pans. Copper has a specific heat of 0.385 . A 25.3 sample of copper heated to 86.0 is dropped into water at an initial temperature of 14.5. If the final temperature reached by the copper-water mixture is 36.0, how many grams of water are present?

Answer 1

c(Cu) = 0.385 J/(g*K)

c(H2O) = 4.186 J/(g*K)

m(Cu) = 25.3
g

t1(Cu) = 86.0 °C

t2(Cu) = 36.0 °C

t1(H2O) = 14.5
°C

t2(H2O) = 36.0 °C

--------------------------

m(H2O) =
?


Solution:


Q = c*m*(t2-t1)

Q(Cu) =
c(Cu)*m(Cu)*(t1(Cu) – t2(Cu)) = 0.385 * 25.3 * (86.0 – 36.0) = 487
J

Q(Cu) = Q(H2O)

Q(H2O) = c(H2O)*m(H2O)*(t2(H2O) – t1(H2O)) =
4.186 * m(H2O) * (36.0 – 14.5)= 487 J

m(H2O) = 5.41
g



Answer:
m(H2O) = 5.41 g