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?

1 Answer
Jul 22, 2016

Answer:

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)#