The specific heat capacity of junior high school physics is 10, and the specific heat capacity of ju

It can be seen from q A = c water m water t, q B = c water m water t.

q A = q B.

And q A = C A M A (t1-t A), Q B = C B M B (t2-t B), i.e.

C A M A (T1-T A) = C B M B (T2-T B) A and B have equal mass, approximate to go, A and B temperature are equal, become.

C A (T1-T A) = C B (T2-T A).

Because the temperature of T1 is higher than that of T2, (T1-TA) > (T2-TA) and therefore C A<>

I have objections. According to the heat calculation formula Q=cmδt, the temperature of water is the same every time it changes, which means that the heat transferred by heat is the same every time.

In this problem, we can understand that after A absorbs heat q, it rises to only 8 from the water temperature, and B absorbs the same amount of heat from the same starting temperature, and is 16 from the initial water temperature, which fully shows that the temperature change of the object is relatively small.

The temperature change of A is the largest, the temperature change of B is small, and the temperature of water is the same each time, which means that the heat transferred by each heat is the same, and the specific heat of B should be greater when the heat is absorbed.

I have an objection and I think the correct answer should be B!

The temperature inference on the first floor has already clearly stated that the temperature change of B is small, so why ignore the formula? I suggest you think about your physics teacher's standard answer to this question!!

The raw water temperature is t, the final temperature of A is t-8, and the final temperature of B is t-16, and the heat released by the two times is equal, so choose A

You have to look at the conditions.

A and B have equal temperatures.

The reason lies in the two words of dropping, which proves that the temperature of A and B is lower than that of water, and the heat released by dropping the temperature of water by 8 degrees is equal, and the temperature of A rises from the original temperature to the temperature of the first cooling.

B's temperature has increased from the original to the temperature of the second cooling, and B has increased less than A's temperature, so you can know it according to q=cmt.

You have to grasp the main conditions and invariants

Select A to substitute separately.

q=cm△t

It came out in the end.

Okay, the definition of specific heat capacity tells us that it is actually a matter of releasing absorbed energy when a substance rises by one degree or decreases by one degree. Let's see if the specific heat capacity of water is the option. Then look at the rate of temperature change with time, who is faster, A, then it means that the specific heat capacity of A should be less than water.

Because the temperature changes quickly, it can accommodate less, which is why the specific heat capacity has the word "capacity" - "the amount of capacity".

And when the temperature changes by 60 degrees, we see if the actual time ratio is 1:2. And according to Q=cmδT, when the mass is the same and the temperature changes the same, c is proportional to the change in heat.

That is to say, the specific heat capacity of a is 1 2Select A

For objects of the same mass, raise the same temperature (both rise to 60) A takes 2 minutes, water takes 4 minutes Water absorbs 2 times the heat of A, according to q=cmδt When the mass and the raised temperature are the same, the greater the specific heat capacity, the more heat absorption, proportional.

Water absorbs 2 times as much heat as A, so the specific heat capacity of AA is 1 2 of water, for.

Q=cmδt C is the specific heat capacity, M is the mass, and ΔT is the temperature difference.

It can be seen from the figure that in the same time, the temperature of A rises twice that of water, and the title says that two identical electric heaters are used to heat A and water of the same mass, indicating that M A = m water, and the heat q absorbed by A and water in the same time is equal, and the formula Q = cmδt can be obtained: C A = 1 2C water = j (kg· ).

Comparing the slope, it can be seen that the specific heat capacity of substance A is less than that of water, the same mass has the same temperature change, and the heating conditions are the same, but the time is half of that of water, so the specific heat capacity is half of that of water.

q=wt, the heating time of water is twice that of A, and the heat absorption of water is twice that of A, so Q water = 2q A, and because q=cm(t-t. ), look at the problem to get the same mass of the two substances, the elevated temperature, so A, so the specific heat capacity of A.

According to Q=CMT, use the same electric heater to heat two substances of the same mass. So q and m in the formula are all the same.

According to the figure: When the temperature of the two substances reaches 60 degrees Celsius (the temperature is changed to the same). The difference in time is about half. The specific heat capacity of water is the answer b, and the time is about half less than that of water, so the specific heat capacity is also option a.

q=cm t can be known to the cubic j(kg.) of q=Celsius) * 1kg * (80-20) = cubic power.

Q = Q water + Q pot = to the third power J(kg.).degrees Celsius) * 2 kg * (80-20) + cubic j (kg.)Celsius) * 1kg * (80-20) = cubic power.

Just remember the formula! Hope it helps!

Iron pot 1kg of the cubic j(kg.Celsius.

Total iron pot + water.

1kg of cubic j(kg.).Celsius + 2 kg to the third power j (kg.).Celsius) think carefully, find the correspondence, and consider it separately, and you can't memorize formulas when learning physics. Hehe.

Heat absorbed by iron q1 = cm (end of t - beginning of t) = = j

Heat absorbed by water Q2 = cm (end of t - beginning of t) = = j

In total, Q total = Q1 + Q2 = J

c.The heating time is the same, and the heat absorbed by liquid A is greater than that absorbed by liquid B.

Amend to: cThe heating time is the same, and the heat absorbed by liquid A is equal to the heat absorbed by liquid B.

That's right. Because the heat absorbed by both liquids comes from the same heat source, that is, the same heat source is heated for the same time, then the heat emitted by the heat source to the two liquids is the same, and the heat absorbed by the two liquids is also the same.

Heating separately with the same heat source means that the heat released at the same time is the same, that is, the two liquids absorb the same amount of heat, so it is equal.

The answer is a:b: the amount of internal energy contained has nothing to do with the specific heat capacity. And the heat cannot be said to be contained, only absorbed or released.

C: The specific heat capacity is only related to the type of substance.

D: It should be emphasized that a substance with a mass of 1 kg ... Can.

A is correct.

b There is no concept of how much heat is contained, and heat can only be said to be exchanged, not how much it contains.

c Specific heat capacity is an intrinsic property of an object that does not change.

d should also be correct.

The specific heat capacity of copper is three times that of lead, copper and iron of equal mass absorb the same heat, and the ratio of their increased temperature is: (1:3), and the mass of B is one-third of that of A, and the temperature of A is twice that of B, then the ratio of heat absorbed by A and B is (6:

1. Just substitute the ratio into the formula.

The ratio of the increase in temperature is: (1:3).

The ratio of heat absorbed is (6:1).

1) M copper: M iron (It's iron!) ) = 3:1, m copper = m iron, q iron = q copper, according to q copper = c copper m copper t q iron = c iron m iron t; Simplified to 1:3.

2) C A = C B, 3m A = M B, T A = 2 t B, according to Q A = C A M A T A Q B = C B M B T B, the ratio is 6:1

The heat capacity of B is larger than A, let the water temperature be t, the temperature heated by A and B is T0, put A into the water and the water temperature rises to T1, and then put B into the water The water temperature is T2. then there are: t1-t=10, t2-t1=10, ca*m*(t0-t1)=cb*m*(t0-t2); Because T2 > T1, CB > CA.

Calorie calculation formula: q=cm t

Q: The amount of heat absorbed (or released).

c: Specific heat capacity (of water).

m: Quality. t: Varying temperature.

It can be seen that q is the same (heat absorbed by water).

In the same way, A first cools down to the same temperature as water, which is set to x (Celsius), and then B raises the water by 10 degrees Celsius, then the temperature of B is 10+x (Celsius).

At the beginning, the temperature of AB is equal, but the temperature of B drops less than that of A, then the heat is calculated according to the formula: Q=CMT

c=q m t, the smaller t, the larger c is. Therefore, the specific heat capacity size b>a

After that, divide by 3.

The above is the amount of heat required to raise 1 degree of lead powder and 2 kg of copper powder After that, divide it by 3 and you can do it, if not, please point out Thank you.

c is the specific heat capacity, which can generally be given in the table or the title, and the specific heat capacity of different substances is different. However, common substances such as water are known by default.