Calculating Specific Heat Capacity: Solving for Unknown Variables

[rrosa522]

Homework Statement

When 1.25 kg of a cold metal at a temperature
of 263 K was immersed in 1.43 kg of water at
a temperature of 365 K, the final temperature
was 336 K. What is the specific heat capacity
of the metal?

Homework Equations

Q=mc∆t
-Q=Q[/B]

The Attempt at a Solution

The answer should be 1.90KJ(kg*K)[/B]

 

[rude man]

Homework Statement

When 1.25 kg of a cold metal at a temperature
of 263 K was immersed in 1.43 kg of water at
a temperature of 365 K, the final temperature
was 336 K. What is the specific heat capacity
of the metal?

Homework Equations

Q=mc∆t
-Q=Q[/B]

The Attempt at a Solution

The answer should be 1.90KJ(kg*K)[/B]

Heat gained by metal = heat lost by water.

 

[CWatters]

The Attempt at a Solution

The answer should be 1.90KJ(kg*K)[/B]

That's not an attempt at a solution.

 

[lep11]

That's not an attempt at a solution.

And there are a couple of typos as well, it should be written as 1.90 kJ/(kgK)

 

[lep11]

Well, you have the right equation there. How much will the temperatures change?