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A 39.3 g glass thermometer reads [tex]22.0^\circ\text{C}[/tex] before it is placed in 136 mL of water. When the water and thermometer come to equilibrium, the thermometer reads [tex]43.6^\circ\text{C}[/tex]. What was the original temperature of the water in [tex]^\circ\text{C}[/tex]?

Do not enter the unit.

Answer :

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:

[tex]E = m_gc_g \Delta T = 39.3 * 0.84 * 21.6 = 713.06 J[/tex]

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

[tex]E = m_wc_w(T - T_w) = 713.06 [/tex]

[tex] 136*4.186(T - 43.6) = 713.06[/tex]

[tex]T - 43.6 = \frac{713.06}{136*4.186} = 1.25[/tex]

[tex]T = 1.25 + 43.6 = 44.85C[/tex]

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