Question

1. Calculate the amount of heat, in kJ, that is transferred to 45.7 mL of water when its temperature changes from 23.6 °C to 41.3 °C. The density of water is 1.00 g/mL and the specific heat of water is 4.184 J/(g•°C).
2. Should the sign of the heat transferred in the question above be positive or negative?

Answer 1

Answer:

from

H=mc∆T

where by;

H is amount of heat.

m is mass of water.

c is specific heat of water.

∆T is change in temperature.

m=density ×volume

m=1.00g/ml×45.7ml

m=45.7g

∆T=Tt-T°

∆T=41.3°C-23.6°C

∆T=17.7°C

H=45.7g×4.184j(g•°C)×17.7°C

H=3,384.396J

H=3,384.396÷1000

H=3.384kJ

Explanation:

H=mc∆T

where by;

H is amount of heat.

m is mass of water.

c is specific heat of water.

∆T is change in temperature