Question

If a cup of tea at 50° C is allowed to cool to room temperature,heat release would be (assume room temperature is equal to 25° C and heat capacity of a cup and tea to be= 5.0 J/K) {Full answer}.

Answer 1

Answer:

When a cup of tea at 50°C is allowed to cool to room temperature, the heat released is 125kJ.

Heat Released:

• Whenever a substance at a higher temperature is kept in a surrounding having a comparatively lower temperature, there is always a flow of heat taking place.
• The heat flow takes place from a body at a higher temperature to a body at a lower temperature.
• The body at a higher temperature is said to release heat whereas the body at a lower temperature is said to take the heat.
• Once the temperature of both the bodies is equal, no further transfer of heat takes place and the bodies are said to be in equilibrium with each other.

Heat Capacity:

• The heat capacity of a substance is a property that states the quantity of heat that is to be provided to the substance so that a unit change in the temperature of the substance is produced.
• It can also be stated as the ratio of heat absorbed by the substance to the change in its temperature.
• It is denoted by C and calculated as $C=\frac{Q}{T_{2}-T_{1} }$, where Q is the amount of heat released and $T_{2} -T_{1}$ is the difference between the two temperatures.
• Unit of heat capacity is joule per degree Celsius or J/°C

Explanation:

Given:

$T_{1} =25$°C, $T_{2}=50$°C, $C=5.0$kJ/°C

To find:

Heat released Q

Formula:

$C=\frac{Q}{T_{2}-T_{1} }$

Step 1: Rearranging the formula

To find the heat released, we multiply the heat capacity by the change in temperature.

$Q=C*(T_{2}-T_{1} )$

Step 2: Substituting and solving

Substituting the values of C, $T_{1}$, and $T_{2}$ in the above formula.

$Q=5.0*10^{3}*(50-25)\\Q=5.0*10^{3}*25\\Q=125*10^{3} J\\Q=125 kJ$

Therefore, the heat released in the cooling of the cup is 125kJ.

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Answer 2

Answer:

The heat released from cup of tea is 125J

Explanation:

Given that,

Initial Temperature of cup of tea=50°C

As it is allowed to cool to room temperature,

The final temperature=25°C

Heat Capacity of cup and tea(H)=5.0J/K

The heat released is given by the relation

$Q=H∆T \\ Q=H( T_{2} - T_{1})$

By substituting given values,we get

$Q=5 \times (50 - 25) = 125J$

Therefore,the heat released from cup of tea is 125J

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