A can is taken out from a refrigerator at 0° C. The atmospheric temperature is 25°C. If t1 is the time taken to heat from 0° C to 5° C and t2 is the time taken from 10° C to 15°C, then :

Question

A can is taken out from a refrigerator at 0° C. The atmospheric temperature is 25°C. If t1 is the time taken to heat from 0° C to 5° C and t2 is the time taken from 10° C to 15°C, then : (A) t1 > t2  (B) t1 < t2  (C) t1 = t2  (D) there is no relation

Tardigrade Admin · Accepted Answer

Sir Newton's law of heating states that rate of change of temperature of a object is proportional to the difference in temperatures of the surrounding and the object. In the given cases, when initial temperature of can is

0^{\circ} C

, the difference in temperatures of can and surrounding is

25 - 0 = 2 5^{\circ} C

.
When initial temperature of can is

1 0^{\circ} C

, the difference in temperatures of can and surrounding is

25 - 10 = 1 5^{\circ} C

therefore the rate of heating will be higher, when temperature difference is

2 5^{\circ} C

i.e. time

t_{1}

will be smaller than

t_{1} < t_{2}

Q. A can is taken out from a refrigerator at $0^{\circ} C$ . The atmospheric temperature is $2 5^{\circ} C$ . If $t_{1}$ is the time taken to heat from $0^{\circ} C$ to $5^{\circ} C$ and $t_{2}$ is the time taken from $1 0^{\circ} C$ to $1 5^{\circ} C$ , then :

$t_{1} > t_{2}$

$t_{1} < t_{2}$

$t_{1} = t_{2}$

there is no relation

Q. A can is taken out from a refrigerator at 0∘C. The atmospheric temperature is 25∘C. If t1​ is the time taken to heat from 0∘C to 5∘C and t2​ is the time taken from 10∘C to 15∘C, then :

t1​>t2​

t1​<t2​

t1​=t2​

there is no relation

Q. A can is taken out from a refrigerator at $0^{\circ} C$ . The atmospheric temperature is $2 5^{\circ} C$ . If $t_{1}$ is the time taken to heat from $0^{\circ} C$ to $5^{\circ} C$ and $t_{2}$ is the time taken from $1 0^{\circ} C$ to $1 5^{\circ} C$ , then :

$t_{1} > t_{2}$

$t_{1} < t_{2}$

$t_{1} = t_{2}$