Question #eca79

1 Answer
Nov 8, 2015

"60 K"60 K

Explanation:

In order to be able to solve this problem, you need to know the specific heat of silver, which is listed as being equal to

c_"silver" = 0.23"J"/("g K")csilver=0.23Jg K

http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html

Now, the idea here is that a substance's specific heat tells you how much energy must be provided in order to increase the temperature of "1 g"1 g of that substance by "1 K"1 K.

In your case, you know that you provide a "32-g"32-g sample of silver with "300 J"300 J worth of heat and want to determine how much will the sample's temperature increase.

The equation that establishes a relationship between heat absorbed and change in temperature looks like this

color(blue)(q = m * c * DeltaT)" "q=mcΔT , where

qq - the amount of heat absorbed
mm - the mass of the sample
cc - the specific heat of the substance
DeltaTΔT - the change in temperature, defined as the final temperature minus the initial temperature of the sample.

So, you need to rearrange this equation and solve for DeltaTΔT

q = m * c * DeltaT implies DeltaT = q/(m * c)q=mcΔTΔT=qmc

Plug in your values to get

DeltaT = (300color(red)(cancel(color(black)("J"))))/(32color(red)(cancel(color(black)("g"))) * 0.23color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) * "K")) = "40.8 K"

This tells you that the temperature of the sample changed by "40.8 K". Therefore, the final temperature will be

DeltaT = T_"final" - T_"initial"

T_"final" = "40.8 K" + "20 K" = "60.8 K"

You need to round this off to one sig fig, the number of sig figs you have for the heat absorbed and for the initial temperature

T_"final" = color(green)("60 K")