Question

Question #5ef67

Answer 1

Since no information is given, I'll provide an example in order to better illustrate what the answer could be.

Here's a nice sample problem (taken from here:http://www.physicsclassroom.com/class/thermalP/Lesson-2/Measuring-the-Quantity-of-Heat):

Suppose we have a 12.9 g sample of an unknown metal at `26.5^@C`. This sample is placed in an insulated cup that contains 50.0 g of water at `88.6^@C`. The water cools down and the metal heats up until thermal equilibrium is reached at `87.1^@C`. Knowing that water's specific heat capacity is 4.18 `"J/"` `g^@C"`, determine the metal's specific heat capacity (assume all the heat lost by the water is gained by the metal).

The heat lost by the water (since it cools down) is equal to

`q = m * c_(H_2O) * DeltaT = 50.0 * 4.18 * (87.1-88.6) = -314J`

The heat gained by the metal (since it heats up) is equal to

`q_(metal) = - q_(H_2O) = m * c_(met a l) * DeltaT`, so

`c_(metal) = (314J)/(12.9 * (87.1-26.5)) = 0.401J/(g * ^@C)`

Notice how much bigger the difference in temperature between the metal and the water was for the same amount of heat gained/lost.

314 J of heat only caused a 1.5-degree drop in temperature for the water, at the same time causing a 60.6-degree increase in temperature for the metal. By comparison, you would need

`q = m * c * DeltaT = 50.0 * 4.18 * 60.6 = 12665J = 12.7kJ` of heat to cause the water to warm up by 60.6 degrees. This is due to the significant difference in specific heat capacity between the two.

Therefore, water is much better at absorbing heat energy (actually, it has the highest specific heat capacity all the liquids) - this is why it's used, for example, in cooling nuclear reactors.