5. Suppose a cup of cocoa cooled from 90° C to 60° C after 10 minutes in a room whose temperature was 20° C . Use Newton’s law of cooling [T − Ts = (T0 − Ts )⋅ e−k⋅t ] to answer the following questions: a) How much longer would it take the cocoa to cool to 35° C ? b) Instead of being left to stand in a room, the cup with initial temperature 90° C is placed in a freezer whose temperature is −15° C . How long will it take the cocoa to cool from 90° C to 35° C ?

a) t = 1051.6 sec = 17.5 min

b) t = 795.5 sec = 13.25 min

Explanation:

First of all we use the initial data to find out constant 'K'.

T - Ts = (T₀ - Ts) e^(-kt)

Here, we have:

T = Final Temperature = 60° C

Ts = Surrounding Temperature = 20° C

T₀ = Initial Temperature = 90° C

t = time = 10 min = 600 sec

k = constant = ?

Therefore,

60° C - 20° C = (90° C - 20° C).e^(-k600)

40° C/70° C = e^(-k600)

ln (0.57142) = -600k

k = 9.327 x 10⁻⁴ sec⁻¹

a)

Now, for this case we have:

T = Final Temperature = 35° C

Ts = Surrounding Temperature = 20° C

T₀ = Initial Temperature = 60° C

t = time = ?

k = constant = 9.327 x 10⁻⁴ sec⁻¹

Therefore,

35° C - 20° C = (60° C - 20° C).e^(-9.327 x 10⁻⁴ sec⁻¹ x t)

15° C/40° C = e^(-9.327 x 10⁻⁴ sec⁻¹ x t)

ln (15/40) = - 9.327 x 10⁻⁴ sec⁻¹ x t

t = 1051.6 sec = 17.5 min

b)

Now, for this case we have:

T = Final Temperature = 35° C

Ts = Surrounding Temperature = -15° C

T₀ = Initial Temperature = 90° C

t = time = ?

k = constant = 9.327 x 10⁻⁴ sec⁻¹

Therefore,

35° C + 15° C = (90° C + 15° C).e^(-9.327 x 10⁻⁴ sec⁻¹ x t)

50° C/105° C = e^(-9.327 x 10⁻⁴ sec⁻¹ x t)

ln (50/105) = - 9.327 x 10⁻⁴ sec⁻¹ x t

t = 795.5 sec = 13.25 min

the answer is b, have negligible mass