Question

a metal sphere cools at the rate of 1.6°C\min when its temperature is 70°C .at what rate will it temperature is 40°C .the temperature of surrounding is 30°C​

Answer 1

Answer:0.4°celcius per min.

Explanation:

According to Newton's law of cooling,

DQ/dt=ha∆t

1.6=ha(70-30)

ha=1.6/40------{eq 1}

Again applying Newton's laws of cooling

DQ/dt=ha*(40-30)

Putting value of ha from eq 1

DQ/dt=[1.6/40]*10

Finally DQ/dt =0.4

Answer 2

Answer:

The new rate of cooling is 0.4°C/min.

Explanation:

We will use the following formula to solve this question,

$R=K(T-T_0)$           (1)

Where,

R=rate of cooling

K=cooling constant

T=given temperature

T₀=tempearature of surrounding

From the question we have,

R=1.6°C/min

T₁=70°C

T₂=40°C

T₀=30°C

We can also modify equation (1) as,

$\frac{R_1}{R_2}=\frac{T_1-T_0}{T_2-T_0}$          (2)

By substituting all the required values in equation (2) we get;

$\frac{1.6}{R_2}=\frac{70-30}{40-30}$

$\frac{1.6}{R_2}=\frac{40}{10}$

$R_2=\frac{1.6}{4}$

$R_2=0.4\textdegree C/min$

Hence, the new rate of cooling is 0.4°C/min.