Question

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?

Answer 1

`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