A 11.1-g sample of granite initially at 76.0°C is immersed into 22.0 g of water initially at 22.0°C. What is the final temperature of both substances when they reach thermal equilibrium? (For water, Cs=4.18J/g⋅∘C and for granite, Cs=0.790J/g⋅∘C.)

Answers

Answer 1

Answer:

[tex]T_f=26.7\°C[/tex]

Explanation:

Hello.

In this case, when two substances at different temperature are placed in contact in an isolated container, we can say that the heat lost by the hot substance is gained by the cold substance. In such a way, since granite is at 76.0 °C and water at 22.0 °C we infer granite is hot and water is cold, so we write:

[tex]Q_{granite}=-Q_{water}[/tex]

In terms of mass, specific heat and change in temperature, we write:

[tex]m_{granite}C_{granite}(T_f-T_{granite})=-m_{water}C_{water}(T_f-T_{water})[/tex]

Thus, since the temperature is the same for both substance, we can solve for it as shown below:

[tex]T_f=\frac{m_{granite}C_{granite}T_{granite}+m_{water}C_{water}T_{water}}{m_{granite}C_{granite}+m_{water}C_{water}}[/tex]

By plugging in each variable, we obtain:

[tex]T_f=\frac{11.1g*0.790\frac{J}{g\°C} *76.0\°C+22.0g*4.18\frac{J}{g\°C} *22.0\°C}{11.1g*0.790\frac{J}{g\°C} +22.0g*4.18\frac{J}{g\°C}}\\\\T_f=26.7\°C[/tex]

Best regards!


Related Questions

The internal energy of reaction is -855.1). The reaction has a change of
temperature of 63.20°C that consist of 8.85g of material. Assume the
heat capacity of 2.650J/g °C. What is the work energy of this process..

Answers

The work energy of this process : 2337.298 J

Further explanation  

The laws of thermodynamics 1 state that: energy can be changed but cannot be destroyed or created  

ΔU=Q-W

Q=m.c.Δt

[tex]\tt Q=8.85\times 2.650\times 63.2=1482.198~J[/tex]

the work (W) :

[tex]\tt W=Q-\Delta U\\\\W=1482.198-(-855.1)=2337.298~J[/tex]

The density of a substance is 1.63 grams per milliliter. What is the volume, in ml, of a sample of the substance with a mass of 5.40Kg? 1000 g = 1 kg

Answers

Answer:

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Explanation:

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