Question

a metal sphere cools @ 0.05 degrees celsius per second when its temperature is 70 degree Celsius and at the rate of 0.0 25 degree Celsius per second when its temperature is 50 degree Celsius determine the temperature of the surroundings and find the rate of cooling when the temperature of metal sphere is 40 degree Celsius​

Answer 1

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Answer 2

The surrounding temperature is 30°C.

The rate of cooling when the temperature of metal sphere is 40 degree Celsius​ is 0.0125°C/s.

Explanation:

Given that,

Rate of cooling $\dfrac{d\theta}{dt}_{1}=0.05^{\circ}C$

Another rate of cooling $\dfrac{d\theta}{dt}_{2}=0.025^{\circ}C$

Temperature $\theta_{1}= 70^{\circ}C$

Another temperature $\theta_{2}= 50^{\circ}C$

We need to calculate the temperature

Using newton's second law

$\dfrac{d\theta}{dt}=k(\theta-\theta_{0})$

For metal sphere at 70° C temperature,

$\dfrac{d\theta}{dt}_{1}=k(\theta_{1}-\theta_{0})$

Put the value into the formula

$0.05=k(70-\theta_{0})$....(I)

For metal sphere at 50°C temperature,

$\dfrac{d\theta}{dt}_{2}=k(\theta_{2}-\theta_{0})$

Put the value into the formula

$0.025=k(50-\theta_{0})$.....(II)

Divided equation (I) by equation (II)

$\dfrac{0.05}{0.025}=\dfrac{(70-\theta_{0})}{50-\theta_{0}}$

$0.05\times50-70\times0.025=0.05\theta_{0}-0.0250\theta_{0}$

$0.75=0.025\theta_{0}$

$\theta_{0}=\dfrac{0.75}{0.025}$

$\theta_{0}=30^{\circ}\ C$

We need to calculate the value of k

Using equation (I)

$k=\dfrac{\dfrac{d\theta}{dt}_{1}}{\theta_{1}-\theta_{0}}$

Put the value in the equation (I)

$k=\dfrac{0.05}{70-30}$

$k=0.00125$

We need to calculate the rate of cooling when the temperature of metal sphere is 40 degree Celsius​

Using formula of rate of cooling

$\dfrac{d\theta}{dt}=k(\theta-\theta_{0})$

Put the value into the formula

$\dfrac{d\theta}{dt}=0.00125(40-30)$

$\dfrac{d\theta}{dt}=0.0125^{\circ} C/sec$

Hence, The surrounding temperature is 30°C.

The rate of cooling when the temperature of metal sphere is 40 degree Celsius​ is 0.0125°C/s.

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Topic : rate of cooling

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