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yawa3891 [41]
3 years ago
14

A 13.5 g sample of gold is heated to 125.0°C, then placed in a calorimeter containing 60.0 g of water. The final temperature of

the water 20.00 o C. The specific heat of gold is 0.130 J/g o C. What was the initial temperature of the water?
Chemistry
1 answer:
jonny [76]3 years ago
3 0

The initial temperature was 19.27 °C.

The guiding principle is the <em>Law of Conservation of Energy</em>: the sum of all the energy transfers must add up to zero.

The formula for the heat <em>q</em> gained or lost by a substance is

<em>q = mC</em>Δ<em>T</em>

where

<em>m</em> = the mass of the substance.

<em>C</em> = its specific heat capacity.

Δ<em>T</em> = <em>T</em>_f - <em>T</em>_i = the change in temperature.

In this problem, there are two heat transfers.

Heat lost by gold + heat gained by water = 0

<em>m</em> _1<em>C</em>_1Δ<em>T</em>_1 + <em>m</em>_2<em>c</em>_2Δ<em>T</em>_2= 0

<em>m</em>_1 = 13.5 g; <em>C</em>_1 = 0.130 J·°C^(-1)g^(-1); Δ<em>T</em>_1 = <em>T</em>_f – <em>T</em>_i = 20.00 °C – 125.0 °C = -105.0 °C

<em>m</em>_2 = 60.0 g; <em>C</em>_2 = 4.184 J·°C^(-1)g^(-1); Δ<em>T</em>_2 = ?

<em>q</em>_1 = <em>m</em>_1<em>C</em>_1ΔT_1 = 13.5 g × 0.130 J·°C^(-1)g^(-1) × -105.0 °C = -184.3 J

<em>q</em>_2 = <em>m</em>_2<em>C</em>_2Δ<em>T</em>_2 = 60.0 g × 4.184 J·°C^(-1)g^(-1) × Δ<em>T</em>_2

= 251.0 ΔT_2 J·°C^(-1)

<em>q</em>_1+ <em>q</em>_2 = -184.3 J + 251.0 Δ<em>T</em>_2 J·°C^(-1) = 0

251.0 Δ<em>T</em>_2 °C^(-1) = 184.3

Δ<em>T</em>_2 = 184.3/251.0 °C^(-1) = 0.734°C

Δ<em>T</em>_2 = <em>T</em>_f - <em>T</em>_i = 20.00 °C -<em>T</em>_i = 0.734 °C

<em>T</em>_i = 20.00 °C – 0.734 °C = 19.27 °C

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