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Valentin [98]
3 years ago
15

PLZZZZ HELP!!!!!

Chemistry
1 answer:
Tpy6a [65]3 years ago
5 0

Answer:

the answer and explanation is in the picture

Explanation:

please like and Mark as brainliest

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A 19.0 g piece of metal is heated to 99.0 °C and then placed in 150 mL of water at 21°C. The temperature of the water rose to 23
Anna35 [415]

Answer:

0.8696\ \text{J/g}^{\circ}\text{C}

Explanation:

m_m = Mass of metal = 19 g

c_m = Specific heat of the metal

\Delta T_m = Temperature difference of the metal = 99-23=76^{\circ}\text{C}

V = Volume of water = 150 mL = 150\ \text{cm}^3

\rho = Density of water = 1\ \text{g/cm}^3

c_w = Specific heat of the water = 4.186 J/g°C

\Delta T_w = Temperature difference of the water = 23-21=2^{\circ}\text{C}

Mass of water

m_w= \rho V\\\Rightarrow m_w=1\times 150\\\Rightarrow m_w=150\ \text{g}

Heat lost will be equal to the heat gained so we get

m_mc_m\Delta T_m=m_wc_w\Delta T_w\\\Rightarrow c_m=\dfrac{m_wc_w\Delta T_w}{m_m\Delta T_m}\\\Rightarrow c_m=\dfrac{150\times 4.186\times 2}{19\times 76}\\\Rightarrow c_m=0.8696\ \text{J/g}^{\circ}\text{C}

The specific heat of the metal is 0.8696\ \text{J/g}^{\circ}\text{C}.

4 0
2 years ago
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