The same amount of heat which will change the temperature of 50.0 g of water by 4.5° C will raise the temperature of 110.0 g of tin from 25° C to 62.7° C. What is the specific heat of tin?

1 Answer
Nov 29, 2015

Answer:

#0.23"J"/("g" ""^@"C")#

Explanation:

The idea here is that if the heat given off by the water is equal to the heat absorbed by the metal, then you can say that

#color(blue)(q_"tin" = -q_"water")#

The minus sign is used to because heat lost carries a negative sign.

Now, the equation that establishes a relationship between heat lost/gained and change in temperature looks like this

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

#q# - heat absorbed/lost
#m# - the mass of the sample
#c# - the specific heat of the substance
#DeltaT# - the change in temperature, defined as final temperature minus initial temperature

The specific heat of water is equal to #4.18"J"/("g" ""^@"C")#.

For water, this equation will take the form

#q_"water" = 50.0 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (-4.5)color(red)(cancel(color(black)(""^@"C")))#

#q_"water" = -"940.5 J"#

For the tin sample, the equation will take the form

#q_"tin" = "110.0 g" * c_"tin" * (62.7 - 25)^@"C"#

This means that you have

#q_"tin" = -q_"water"#

#q_"tin" = - (-"940.5 J") = +"940.5 J"#

Therefore,

#"940.5 J" = "110.0 g" * c_"tin" * 37.7^@"C"#

#c_"tin" = "940.5 J"/("110.0 g" * 37.7^@"C") = 0.2267"J"/("g" ""^@"C")#

You need to round this off to two sig figs, the number of sig figs you have for the change in temperature of the water

#c_"tin" = color(green)(0.23"J"/("g" ""^@"C"))#

The accepted value for tin's specific heat is #0.21"J"/("g" ""^@"C")#, so your result is fairly accurate.

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