Water metal heat transfer?

Question

if you used metal with a specific heat twice that of silver, leaving everything else the same, what would the change in temperature of the water be?

If you used silver with half the mass leaving everything else the same, what would the change in temperature of the water be?

I don't need to give specific numbers just say it would be half as warm or something

Also how do u determine the specific heat for an unknown metal? Like without knowing E or c.
E=mc^T (The ^ stands for a triangle)

Answer 1

Looking this up has been interesting. A number of sites give the specific heat of silver as 0.235 J/g-°C, which is its heat capacity. The specific heat of silver is then 0.235/4.184 = 0.056166 since the specific heat of water is 1 by definition.

Using upper case for variables and lower case for subscripts, the energy balance equation for silver would be
0.056166Ms∆Ts = Mw∆Tw
Solving for ∆Tw,
∆Tw = 0.056166Ms∆Ts/Mw
For the unknown metal,
∆Tw = 2*0.056166Ms∆Ts/Mw, or twice what it would be for silver.

Using silver with half the original mass and everything else the same, ∆Tw would be half the original ∆Tw.

To determine the specific heat of an unknown metal, let Sm = specific heat of the metal,
SmMm∆Tm = Mw∆Tw
Sm = Mw∆Tw/Mm∆Tm

Answer 2

What was the original question.