Question
A 37.0 g mass of a metal was heated to 100°C and then plunged into 74 g of water at 24.0°C. The temperature of the resulting mixture became 28.0°C.
(a) How many joules did the water absorb?
(b) How many joules did the metal lose?
(c) What is the heat capacity of the metal sample?
(d) What is the specific heat of the metal?
(a) How many joules did the water absorb?
(b) How many joules did the metal lose?
(c) What is the heat capacity of the metal sample?
(d) What is the specific heat of the metal?
Answers
a.
joules H2O absorbed = mass H2O x spcific heat H2O x (Tfinal-Tinitial)
b.
joules absorbed by H2O = heat gained by metal
d.
q metal = mass metal x specific heat metal x (Tfinal-Tinitial)
c. q = mass metal x delta T
joules H2O absorbed = mass H2O x spcific heat H2O x (Tfinal-Tinitial)
b.
joules absorbed by H2O = heat gained by metal
d.
q metal = mass metal x specific heat metal x (Tfinal-Tinitial)
c. q = mass metal x delta T
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