A 50.0 g piece of metal at 100.0 degC is dropped into 100.0 g of water at 20.0 degC in an insulated container; the final temperature is 26.0 degC. Take c(water) = 4.18 J g^-1K^-1 and ignore the heat capacity of the container. The specific heat of the metal in J g^-1K^-1 (three decimals) is

JEE Main levelNumerical answerDifficulty 2 of 3td027

A $50.0 g$ piece of metal at $100.0^{\circ} C$ is dropped into $100.0 g$ of water at $20.0^{\circ} C$ in an insulated container; the final temperature is $26.0^{\circ} C$ . Take $c (water) = 4.18 J g^{- 1} K^{- 1}$ and ignore the heat capacity of the container. The specific heat of the metal in $J g^{- 1} K^{- 1}$ (three decimals) is

Numerical answer type, answer in J/(g K), tolerance 0.002.

Heat lost by metal = heat gained by water:

50.0 c (100.0 - 26.0) = 100.0 (4.18) (6.0)

. So

c = 2508/ (50.0 \times 74.0) = 0.6778 J g^{- 1} K^{- 1}