Help with chemistry heat transfer question?

Question

how much will the temperature of a cup (130 g) of water at 87 degrees celsius be reduced when a 34-g silver spoon (with a specific heat of 0.24 J/g degrees celsius) at 23 degrees celsius is placed in the coffee and the two are allowed to reach the same temperature?

this is what i've done:

q=(34 g)(0.24 J/gC)(x-25 degrees celsius)

q=(130 g)(4.184 J/gC)(x-95 degrees celsius)

is this how you set up the problem?

i'm totally confusing myself :(

Answer 1

You have the right idea that q water = q spoon. What you have wrong is the temperature change. The water is going to cool down, so its delta T is its initial temperature minus the final temperature Tf (i.e.,87-Tf). The spoon is going to absorb heat from the water so its delta T will be the final temperature minus the initial temperature (Tf-23).

So (34)(0.24)(Tf-23) = (130)(4.184)(87-Tf)

Solve for Tf and subtract from 87 to find the final temperature of the water.

(I get Tf = 86 C)

Answer 2

Heat Gained = Heat Lost.

'T' = final temp. °C.

Water loses. 130g x 4.184 x (87°C - T).

= 47,321 - 544T.

Silver gains. 34g x 0.24J/g/°C x (T - 23°C).

= 8.2T - 188.

47,321 + 188 = 544T + 8.2T

= 47,509 = 552.2T

T = 47,509 / 552.2 = 86°C Final temp.

Answer 3

You are setting it up correctly.

Heat lost = heat gained.

Q = m s ΔT

Q = Q

You are looking for ΔT of the water.

For the spoon:

34g x 0.24J/gºC x (87-23)ºC

Q =  544J

For the water:

554J = 130g x 4.184J/gºC x Δ?ºC

Δ?º = 554 / 130 x 4.184

= 1.0ºC

87º - 1.0ºC = 86ºC