What is the specific heat of a metal if 298.0 grams of the metal causes 300.00 grams of water to increase in temperature from 35.0°C to 48.8°C? (The original temperature of the metal was 95.6°C; its final temperature was 48.8°C.)​

Suppose a piece of iron with a mass of 21.5 g at a temp of 100.0 °C is dropped into an insulated container of water. The mass of the water is 132.0 g and its temperature before adding the iron is 20.0 °C. What will be the final temp of the system? Specific heat of iron is 0.449 kJ/kg K.

Solution:

1) Since

qlost, metal = qgained, water

we write

(mass) (Δt) (Cp, metal) = (mass) (Δt) (Cp, water)

2) Substituting:

(21.5) (100 − x) (0.449) = (132.0) (x − 20) (4.184)

Some explanation:

a) 100 − x is the Δt for the metal; it starts at 100.0 °C and drops to some unknown, final value.

b) x − 20 is the Δt for the water; it starts at 20.0 °C and rises to some unknown, final value.

c) Since both metal and water wind up at the same ending value, we need to use only one unknown for the two Δt expressions.

3) A wee bit of algebra:

(2150 − 21.5x) (0.449) = (132x − 2640) (4.184)

965.35 − 9.6535x = 552.288x − 11045.76

561.9415x = 12011.11

To 3 sig figs, the answer is 21.4 °C.

Explanation:

hope its correct!