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2 years ago

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A piece of glass with a mass of 32.50 g specific heat of 0.840 J/g*°C and an initial temperature of 75 °C was dropped into a cal

orimeter containing 57 g of water (specific heat 4.184 J/g*°C). The final temperature of the glass and water in the calorimeter was 79.2 °C. What was the initial temperature of the water?

2 answers:

elixir [45]2 years ago

8 0

Hey there,

119.84 degrees c is your answer.

Hope I Helped!

Have a great day!

DanielleElmas [232]2 years ago

3 0

Answer : The initial temperature of water is, $79.7^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of glass = $0.840J/g^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of glass = 32.50 g

$m_2$ = mass of water = 57 g

$T_f$ = final temperature of mixture = $79.2^oC$

$T_1$ = initial temperature of glass = $75^oC$

$T_2$ = initial temperature of water = ?

Now put all the given values in the above formula, we get:

$32.50g\times 0.840J/g^oC\times (79.2-75)^oC=-57g\times 4.184J/g^oC\times (79.2-T_2)^oC$

$T_2=79.7^oC$

Therefore, the initial temperature of water is, $79.7^oC$

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Hello,

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