Question

Question #1dcd8

Answer 1

`"0.1457 J g"^(-1)""^@"C"^(-1)`

Explanation:

The idea here is that the heat lost by the metal will be equal to the heat absorbed by the water.

`color(blue)(-q_"metal" = q_"water")`

Here the minus sign is used because, by definition, heat lost carries a negative sign.

Your tool of choice here will be the equation

`color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "`, where

`q` - the amount of heat gained / lost
`m` - the mass of the sample
`c` - the specific heat of the substance
`DeltaT` - the change in temperature, defined as the difference between the final temperature and the initial temperature

You know the mass of water, its initial temperature, the final temperature of the system, and water's specific heat, which means that you can determine how much heat was needed to increase the temperature of the water by

`DeltaT_"water" = 16.19^@"C" - 14.00^@"C" = 2.19^@"C"`

Plug in your values to get

`q_"water" = 170.0color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 2.19color(red)(cancel(color(black)(""^@"C")))`

`q_"water" = "1556.214 J"`

So, this is how much heat was absorbed by the water, which can only mean that this is how much heat was lost by the metal.

You can thus say that

`-q_"metal" = "1556.214 J"`

The change in temperature for the metal will be

`DeltaT_"metal" = 16.19^@"C" - 79.00^@"C" = -62.81^@"C"`

This means that you have

`color(red)(cancel(color(black)(-)))"1556.214 J" = "170.0 g" * c_"metal" * (color(red)(cancel(color(black)(-)))62.81)^@"C"`

The specific heat of the metal will be

`c_"metal" = "1556.214 J"/("170.0 g" * 62.81^@"C")`

`c_"metal" = "0.14574 J g"^(-1)""^@"C"^(-1)`

Rounded to four sig figs, the answer will be

`c_"metal" = color(green)(|bar(ul(color(white)(a/a)"0.1457 J g"^(-1)""^@"C"^(-1)color(white)(a/a)|)))`