Mass of water in container = .0231 g

[pmahesh107]

A sample of silver with a mass of 63.3 g is heated to a temperature of 384.4 K and placed in a container of water at 290.0 K. The final temperature of the silver and water is 292.4 K. Assuming no heat loss, what mass of water was in the container? The specific heat of water is 4.184 J/(g C) and of silver, 0.24 J/(g C).

I don't understand what to do for this problem. When I set it up, I end up having 2 unkowns for water. Can someone please show me how to set this up? thanks

i tried it a number of different ways but the only way i could get mass was


4.184 = x/ 181.15 <- converted to celsius from Kelvin

4.184/181.15 = x

.0231 g = x

 

[mrjeffy321]

Remeber two things,
first, the heat lost by the Silver must equal the heat gained by the water, and the heat needed to change the temperature of something equals,
Q = mC(delta T)

We are given the mass, specific heat, and (indirectly), the change in temperature of the Silver, so we can easily calculate the heat energy lost to the water by plugging into the formula.

We are given the specific heat of water as well as indirectly, the change in temperature (final - initial temperatures). We know that all the heat energy the Silver lost, the water absorbed.
Setting the energy lost by the silver (call it Q1) equal to the energy gained by the water (call it Q2) using the formula,
Q1 = Q2 = (mass of the water)*(specific heat of water)*(delta T of water)
Just plug in and solve for the mass of the water.

 

[pmahesh107]

thanks. i forgot that the Q value for silver can be plugged in for water.