Question

A coffee-cup calorimeter contains 130.0 g of water at 25.3 ∘C . A 124.0-g block of copper metal is heated to 100.4 ∘C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g⋅K . The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.3 ∘C .

Part A

Determine the amount of heat, in J , lost by the copper block. (3350 J)

Part B

Determine the amount of heat gained by the water. The specific heat of water is 4.18 J/g⋅K . (2720 J)

Part C

The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam® cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by 1 K . Calculate the heat capacity of the calorimeter in J/K .

Express your answer using two significant figures.

Part D

What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?

Answer 1

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (kg)

c = specific heat (units J/kg∙K), ∆ is a symbol meaning "the change in"

∆T = change in temperature (Kelvins, K)

Part A

heat, in J , lost by the copper block

Q = 124g x 0.385 J/g⋅K x ( 100.4 ∘C -30.3 ∘C)

Q =3346.574 Joules  

Part B

The amount of heat gained by the water

Q = 130g x 4.184 J/g⋅K x ( 30.3 ∘C -25.3  ∘C)

Q =2719.6Joules  

Part C

heat capacity of the calorimeter   3346.574 Joules - 2719.6 Joules / 5 ∘C =  125.3948 ∘C

we can use either ∘C or kelvin

Part D

Heat lost by copper = Heat gained by water

124g x 0.385 J/g⋅K x ( 100.4 ∘C - X ∘C) =  130g x 4.184 J/g⋅K x ( X -25.3  ∘C)

4793.096 - 47.74 X = 543.92 X -13761.176  

18554.272 = 591.66 X

X = 31.359 ∘C

Hence final temperature will be  31.359 ∘C