Question

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

Answer 1

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

Explanation:

`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`