How do I solve this problem

Question

A 90.0 g piece of metal, initially at 98.6 Celsius, is placed into 120.0 g of water initially at 24.3 Celsius. The final temperature of the water is 34.0 Celsius what is the specific heat of metal? (that’s specific heat of water is 4.18 j/g Celsius)

Answer 1

Hey Yesenia!

To do this problem, we have to set up two equations for q=mc∆t. We can express this as two equivalent q's (but opposites in sign), as the metal is put into water, and one increases in temperature (water) while the other decreases in temperature (metal). Doing so, we get:

q=mc∆t (water) and q=mc∆t (metal), where both values for q are equivalent. However, the q for water will be positive, as its temperature increases, while the q for the metal is negative, as its temperature decreases.

q=mc∆t (water)

q= (120g)(4.184J/gC)(34C-24.3C)

q= (120g)(4.184J/gC)(9.7)

q= 4870.18J (water)

We now know that water has a q of 4870.18J, so this means that the metal that was put into the water lost 4870.18J of energy, which means that q for the metal will be -4870.18J. The final temperature of the metal will be the same final temperature of the water, 34.0C as it is submerged in the water. Now, solving q=mc∆t for the metal, we get:

q=mc∆t

q/(m∆t)=c

-4870.18J/(90.0g * (34.0C-98.6C))

-4870.18J/(90.0g *-64.6C)

c= 0.838J/gC

Hope this helps!