Question

Question #66cc0

Answer 1

The specific heat capacity of the metal is `"0.32 J°C"^"-1""g"^"-1"`.

Explanation:

There are two heat transfers involved.

`"heat lost by the metal + heat gained by the water = 0"`

`q_1 + q_2 = 0`

`m_1C_1ΔT_1 + m_2C_2ΔT_2 = 0`

In this problem,

`m_1 = "16.5 g"`; `C_1 = ?`; `color(white)(mmmmm)ΔT_1 = T_"f" - T_"i" = "7.5 °C - 93.0 °C" = "-85.5 °C"`
`m_2 = "84.0 g"`; `C_2 = "4.18 J°C"^"-1""g"^"-1"`; `ΔT_2 = T_"f" - T_"i" = "7.5 °C - 6.2 °C" = "1.3 °C"`

`q_1 = m_1C_1ΔT_1= "16.5 g" × C_1 × ("-85.5 °C") = "-1411"C_1 color(white)(l)"g°C"`

`q_2 = m_2cΔT = 84.0 color(red)(cancel(color(black)("g"))) × "4.18 J"color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × 1.3 color(red)(cancel(color(black)("°C"))) = "456 J"`

`q_1 + q_2 = "-1411"C_1 color(white)(l)"g°C" + "456 J" = 0`

`C_1 = "456 J"/(1411color(white)(l)"g°C") = "0.32 J°C"^"-1""g"^"-1"`.