Question

A 17.5 g sample of metal at 125.0°C is placed in a calorimeter with 15.0 g of water at 25.0°C. if the temperature of the water rises to 30.0°C, what is the specific heat of the metal?

Answer 1

The specific heat of the meta is `C_p = 0.189 J/(g^oC)`

Explanation:

This is a thermo-equillibrium situation. We can use the equation

Loss of Heat of the Metal = Gain of Heat by the Water

`-Q_(m) = +Q_(w)`

`Q = mDeltaTC_p`

`Q =` Heat
`m =` mass
`DeltaT = (T_f-T_i)`
`T_f=` Final Temp
`T_i=` Initial Temp
`C_P=` Specific Heat

Metal
Water
`m =17.5 g`
`T_f=30.0^oC`
`T_i=125.0^oC`
`C_P= x`

Water
`m =15.0 g`
`T_f=30.0^oC`
`T_i=25.0^oC`
`C_P= 4.184 J/(g^oC)`#

`-Q_(m) = +Q_(w)`
`-[m(T_f-T-i)C_p] = m(T_f-T-i)C_p`

`-[17.5g(30^oC-125^oC)x] = 15g(30^oC-25^oC)4.184J/(g^oC)`

`-[17.5g(-95^oC)x] = 15cancel(g)(5^ocancelC))4.184J/(cancel(g^oC))`

`(1662.5g^oC)x = 313.8J`

`cancel(1662.5g^oC)x/cancel(1662.5g^oC) = (313.8J)/(1662.5g^oC)`

`C_p = 0.189 J/(g^oC)`