Question

When 71.0 g of a metal at 112.8 oC are added to a calorimeter containing 160.8 g of water at 15.5 oC, the temparature of the water rises to 21.1 o C.

The specific heat of water is 4.184 J/goC.

What is the specific heat of the metal? Assume no heat is lost to the calorimeter or the surroundings.

Answer 1

@lila2015 , where exactly you are facing a problem??

Just remember two things for these type of sums
1)
\[Q = M*c*\Delta t\]
Q-->energy required
ΔT -->Change in temperature
C-->Specific heat
M-->mass of the object
and during transition from solid ->liquid state , use this formula

2)
\[Q=M*L\]
L-->Latent heat of fusion/vapourisation (depending on condition)

3)
Calorimetry principle

Heat lost by one = Heat gained by other