What is the calories needed to heat 11.4 g of water from 21.0°C to 43.0°C?

Question

1995 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-12T02:30:23

You will need to use the equation below to solve this problem.

$Q = mcΔT$

m - mass in grams of the substance

c - specific heat capacity (J/g°C) (Varies according to the state of substance)

ΔT - Change in temperature (°C)

Here, it is given that the mass of water is 11.4 grams. Meanwhile, the change in temperature would be

$43.0°C - 21.0°C = 22.0°C$

According to my textbook, the specific heat capacity for liquid water is 4.19J/g°C

Substitute all of these values into the equation and we will obtained

$Q = 11.4g * 22.0°C * 4.19J/g°C$

$Q = 1050.8 J$

If we were to convert it into kJ and take significant figures into consideration, the answer would be

$1.05 kJ$

As 2,500kcal is equivalent to 10,500 kJ

$(1.05 kJ * 2,500kcal)/10500kJ)$

Final Answer $0.25kcal$