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

[mass water x specific heat water x (Tfinal-Tinitial)] + [mass Ag x specific heat Ag x (Tfinal-Tinitial) = 0

Solve for Tfinal.

139.8

Well, this is quite a heaty situation you've got here! Let me put on my thinking cap and help you out.

To solve this problem, we can use the principle of heat transfer, which states that the heat gained by the water is equal to the heat lost by the silver. So, we have:

Heat gained by water = Heat lost by silver

The heat gained by the water can be calculated using the formula Q = m * c * ΔT, where Q is the heat gained, m is the mass of water, c is the specific heat of water, and ΔT is the change in temperature.

The heat lost by the silver can also be calculated using the same formula, with the mass and specific heat of silver.

So, we have two equations:

m_water * c_water * ΔT_water = m_silver * c_silver * ΔT_silver

Given that the final temperature of both the silver and water is 292.4 K, we can calculate the change in temperature for both substances:

ΔT_water = 292.4 K - 290.0 K = 2.4 K
ΔT_silver = 384.4 K - 292.4 K = 92 K

Now we can plug in the known values and solve for the mass of water:

m_water * 4.184 J/(g.K) * 2.4 K = 63.3 g * 0.24 J/(g.K) * 92 K

Simplifying, we get:

m_water = (63.3 g * 0.24 J/(g.K) * 92 K) / (4.184 J/(g.K) * 2.4 K)

Now, let me do the calculation for you:

m_water ≈ 3.64 g

So, the mass of water in the container is approximately 3.64 grams.

I hope I was able to "heat" up your understanding of this problem a little! If you have any more questions, feel free to ask.

To solve this problem, we can apply the principle of heat transfer:

Q(silver) + Q(water) = 0

First, let's calculate the heat gained by the silver:

Q(silver) = m(silver) * C(silver) * ΔT(silver)

where:
- Q(silver) is the heat gained by the silver,
- m(silver) is the mass of the silver,
- C(silver) is the specific heat of silver,
- ΔT(silver) is the change in temperature of the silver.

Substituting the given values:
Q(silver) = 63.3g * 0.24J/(g.C) * (292.4K - 384.4K)

Next, let's calculate the heat lost by the water:

Q(water) = m(water) * C(water) * ΔT(water)

where:
- Q(water) is the heat lost by the water,
- m(water) is the mass of the water (which we need to find),
- C(water) is the specific heat of water,
- ΔT(water) is the change in temperature of the water.

We know that the final temperature of the silver and water is 292.4K, so the change in temperature of the water, ΔT(water), is:

ΔT(water) = 292.4K - 290.0K

Now, we can rewrite the equation for heat transfer to solve for the mass of the water:

Q(water) = -Q(silver)

m(water) * C(water) * ΔT(water) = -Q(silver)

Substituting the values we have, we can solve for m(water):

m(water) = -Q(silver) / (C(water) * ΔT(water))

Now, plug in the values and calculate:

m(water) = -Q(silver) / (4.184J/(g.C) * (292.4K - 290.0K))

This will give you the mass of water in grams that was in the container.