How do i solve this?

Question

A 12.9g sample of a metal is heated to 100.^o C and used to heat 46.5g of water from 23.0^o C to 35.0^o C, the specific heat of the metal is ______ J/g^o C. 

Answer 1

I don't believe the previous answer is correct. Let me show you how I approach this probem.

Heat lost by the metal must equal heat gained by the water.

q = mC∆T

q = heat

m = mass

C = specific heat

∆T = change in temperature

For the heat lost by the metal we have ...

q = (12.9 g)(C)(100 - 35) note: the change in temperature is from 100º to 35º, the final temperature.

q = (12.9 g)(C)(65º)

For the heat gained by the water we have ...

q = (46.5 g)(4.184 J/gº)(35 - 23º)

q = (46.5 g)(4.184 J/gº)(12º)

Setting these equal we have ...

(12.9 g)(C)(65º) = (46.5 g)(4.184 J/gº)(12º)

C = (46.5 g)(4.184 J/gº)(12º) / (12.9 g)(65º)

C = 2.78 J/gº

Answer 2

The equation you need to solve this problem is: Q = Δ T * m * s

where Q = the amount of heat added, Δ T = change in temperature, m = grams, s = specific heat

First, use this equation to find Q for water

Δ T = 35.0°C - 23.0°C = 12.0°C

m = 46.5g

s = 4.184 J/g•°C [specific heat of water; usually provided on an equation sheet or should be memorized]

Q_water= 12.0°C * 46.5g * 4.184 J/g•°C = 2,334.672 J

(The units cancel out so just J remains. I usually wait until the end to round to the appropriate amount of significant figures.)

Q_water = Q_metal because the heat lost by the metal is equal to the heat gained by the water

Δ T = 100°C - 35°C = 65°C

m = 12.9g

Now, use the same equation to find s for the metal

Q_metal = 2,334.672 J = 65.0°C * 12.9g * s

s = 2.78 J/g•°C

Let me know if you need any clarification of have further questions :)