Answer to Question #266560 in General Chemistry for Micheal

Question

Part A

50 grams of ice-cold water (0 °C) is heated to boiling (100 °C). The specific heat capacity of water is 4.18 J /g°C. How much energy did the water absorb?

Part B

Over the next hour, the water gradually cools to 25 °C. How much energy did the water lose?

Part C

What was the temperature of the room that the water was in?

Answer 1

Part A

$Q = mcΔT \\ m= 50\; g \\ ΔT = 100 -0 = 100 \;°C \\ Q = 50 \times 4.18 \times 100 = 20900 \;J$

Part B

$ΔT = 25-100 = -75 \; °C \\ Q = 50 \times 4.18 \times (-75) = -15675 \; J$

the water lose 15675 J

Part C

The temperature of the room was 25 °C