Question

A 150g sample of mercury and150g sample of iron are an initial temperature of 25.0c if 250 cal (1050 joul) of heat is applied to each sample. What is the final temperature of each?

Answer 1

Answer: 75° C for mercury, 40.59° C for iron

Step-by-step explanation:

the equation used to solve this problem is q=mcΔT ( where q is heat in joules, m is mass in grams, c is the specific heat capacity, and ΔT is the change in temperature in degrees celsius)

For mercury:

q = mcΔT

1050J = (150g)(0.140J/g°C)(final T - 25°C)

`(1050)/((150)(0.140))` = final T - 25

Final T =
`(1050)/((150)(0.140))` - 25

Final T = 75°C

For iron:

q = mcΔT

1050J = (150g)(0.449J/g°C)(final T - 25°C)

`(1050)/((150)(0.449))` = final T - 25°C

final T =
`(1050)/((150)(0.449))` + 25

Final T = 40.59°C