What quantity of heat is required to raise the temperature of 0.1 kg of water by 1⁰c

0.418 Joules

Explanation:

Recall that the formula :

Heat Added = Specific heat x mass x change in temperature

or

Q = mcΔT

in this case we are given that m = 0.1 kg, ΔT = 1°C  (aka 1 kelvin).

we also know that c for water is 4.1813 J/kg-K

substituting this into the formula

Q = 0.1 x 4.1813 x 1 = 0.41813 = 0.418 Joules ( 3 dec pl)

v=ir = 3x15 =45 v

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s