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adell [148]
3 years ago
13

A 39.3 g glass thermometer reads 22.0oC before it

Physics
1 answer:
ratelena [41]3 years ago
6 0

Answer:

44.85C

Explanation:

Let the specific heat of glass thermometer be 0.84 J/g°C

Let the specific heat of water be 4.186 j/g °C

Let the water density be 1kg/L

136 mL of water = 0.136L of water = 0.136 kg of water = 136 g of water

Since the change of temperature on the glass thermometer is 43.6 - 22 = 21.6 C. We can then calculate the heat energy absorbed to it:

E = m_gc_g \Delta T = 39.3 * 0.84 * 21.6 = 713.06 J

Assume no energy is lost to outside, by the law of energy conservation, this heat energy would come from water

E = m_wc_w(T - T_w) = 713.06

136*4.186(T - 43.6) = 713.06

T - 43.6 = \frac{713.06}{136*4.186} = 1.25

T = 1.25 + 43.6 = 44.85C

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Answer:

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Explanation:

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Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

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ρ = density

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