If a substance has a high specific heat, is it easy or difficult to change the temperature of that

Physics

1 answer:

zmey [24]4 years ago

7 0

It's difficult because the temperature is obviously maintained well if it's designated a specific heat.

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Is block b speeding up or slowing down once it is set into downward motion and the cat is on block a?

stepan [7]

Answer:

The answer is "Slowing down ".

Explanation:

please find the complete question in the attached file.

In this question, if the block B weight were accounted for by kinetic the friction of frame A, because the blocks pushed at a consistent speed throughout the beginning.

Afterward, on block A, the resistance intensity rises, which allows frames to also be negative, which is defined in the graph, that's why the answer Slowing down is correct.

6 0

3 years ago

A 12 oz can of soda was left in a car on a hot day. in the morning, the soda temperature was 60°f with a gauge pressure of 40 ps

enot [183]

From gas equations
PV/T = Constant

Then
P1V1/T1 = P2V2/T2, but V1 = V2

Therefore,
P1/T1 = P2/T2
P2 = (P1T2)/T1

P1 = 40 psi
T1 = 60°F = 288.706 K
T2 = 90°F = 305.372 K

Substituting;
P2 = (40*305.372)/288.706 = 42.31 psi

Then, gauge pressure would read 42.31 psi.

4 0

4 years ago

A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops

Ilya [14]

Answer:

(a) Compression work at the final state with a pressure of 1(MPa) is: 44.32(KJ), (b) Compression work at the final state with a pressure of 500(KPa): 110.37(KJ) and (c) temperaure of the final state in part b: T=151.83(°C).

Explanation:

Remember that the substance is steam so it's water (H2O) and the initial conditions are $P_{1} =1MPa, T_{1}=400^{0}C, m=0.6Kg$ and $v_{2} =0.4v_{1}$ from a saturated water table and the initial conditions we can determine that the state phase is superheated (see Table 1 attached) because the $T_{sat}=179.88^{0} C \leq T_{1}$ from the table 1 we get: $v_{1} =0.30661(m^{3}/Kg)$ . Now we have second conditions as: $P_{2}=1(MPa), T_{2}=250^{0}C$ so from the same table we can see the state still superheated and we get $v_{2}=0.23275(m^{3}/Kg)$ , knowing that it's a isobaric process we can find the compression's work as: $W_{b}=m*P(v_{2}-v_{1})=0.6*1000*(0.23275-0.30661)=-44.32(KJ)$ so the compressor's work is: 44.32(KJ). (b) Then the piston reaches the stop and there are two processes in this stage, so Process 1 is isobaric and: $W_{1}=m*P*(v_{2}-v_{1}) =0.6*1000*(0.4*0.30661-0.30661)=-110.38(KJ)$ and the second process is isochoric: $W_{2}=zero$ ,now $W_{b}=W_{1}+ W_{2} =110.38+0=110.38(KJ)$ . Finally to get the temperarure at the final state in part (b) we get: $v_{2} =0.4v_{1} =0.4*0.30661=0.122644(m^{3}/Kg), P_{2}=500(KPa)$ from table 2 (see attached) we compare $v_{f}$ and $v_{g}$ at the saturated water table and find the following: $v_{f}=0.001093(m^{3}/Kg)$ , so we know that the final state phase is a satured mixture and we get the temperature at the final state as: $T_{2} =T_{sat} =151.83^{0}C$ .