He heats 19.6 grams of gold to 98.44 °C and then drops it into an insulated cup containing 83.0 grams of water at 20.91 °C.

Question

adding the last part of the question here, could not fit in title.

When thermal equilibrium is reached, he measures the final temperature to be 21.46 °C.

Assuming that all of the heat is transferred to the water, he calculates the specific heat of gold to be  J/g°C.

Answer 1

As with your other questions, use q = mC∆T

q = heat

m = mass

C = specific heat

∆T = change in temperature

Heat lost by hot gold must equal heat gained by cooler water

heat lost by gold = q = mC∆T

q = (19.6 g)(C)(98.44 - 21.46) = 1509 gº x C

heat gained by water = q = mC∆T

q = (83.0 g)(4.184 J/gº)(21.46 - 20.91) = 191 J

Setting the two equal to each other we have ...

1509 gº x C = 191 J

C = 0.127 J/gº