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S_A_V [24]
1 year ago
6

What is the temperature change of an 36-gram sample of water that had absorbed 695 Joules of heat? C = 4.18 J/g K

Chemistry
1 answer:
maks197457 [2]1 year ago
6 0

The temperature change of a 36-gram sample of water is <u>4.61 K</u>

Temperature is the degree of the average kinetic energy of the debris in an item. while the temperature will increase, the movement of this debris additionally will increase. Temperature is measured with a thermometer or a calorimeter.

heat = msΔt

     695 Joule = 36 ×  4.186 × Δt

Δt   =  695/ 36 ×  4.186

      = <u>4.61 K</u>

Temperature is just a dimension that indicates the common kinetic energy of 1 atom or molecule. as a result, when we are saying something is warm or bloodless, we're generally using any other reference point to define the hotness and coldness of a frame.

Learn more about temperature here:-brainly.com/question/2339046

#SPJ9

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hope this helps!


4 0
3 years ago
3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,
VARVARA [1.3K]

Answer:

Cp_{liquid}=2.54\frac{J}{g\°C}

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

Q_{Ag}=-Q_{liquid}

That in terms of the heat capacities, masses and temperature changes turns out:

m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}

Best regards.

6 0
4 years ago
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