7. If 294 J of heat is transferred to a 10.0 g sample of silver at 25 ºC, what is the final temperature of the silver? Specific heat capacity of silver is 0.235 J/gºC.

T~final is 150.11 C

Step-by-step explanation:

To solve this we need this specific heat equation:

Q=mc(T~final - T~initial) that is the difference of final temperature minus initial temperature.

We are given:

Q = 294 J

m = 10 g

c = 0.235 J/gC

T~initial = 25 C

Plug the values in to the equation

294 J = 10 g * 0.235 J/gC * T(difference)

294 J = 2.35 J/C * T(difference)

Divide both sides by 2.35 J/C

294 J / 2.35 J/C = T(difference)

T(difference) = 125.11 C

If the initial temperature was 25 C then then

T~final = 25 C + 125.11 C or 150.11 C

the f

last graph i believe

step-by-step explanation:

worked it out i believe

i cant answer this, theres no table. sorry man

step-by-step explanation: