Question

If 425.0 g of mercury at 25.0 degrees Celsius absorbs .750 Kilo joules of heat energy, what will be it's final temperature?

Answer 1

`"37.6°C"`

Explanation:

Use this equation

`"Q = mSΔT"`

Where

• `"Q ="` Heat absorbed/released
• `"m ="`Mass of sample
• `"S ="` Specific heat of sample (`bb"0.14 J/g°C"` for mercury)
• `"ΔT ="` Change in temperature of sample

`"ΔT" = "Q"/"mS"`

`"T"_f - "T"_i = "Q"/"mS"`

`"T"_f = "T"_i + "Q"/"mS"`

`color(white)("T"_f) = "25.0°C" + "750 J"/"425.0 g × 0.14 J/g°C"`

`color(white)("T"_f) = "25.0°C + 12.6°C"`

`color(white)("T"_f) = "37.6°C"`