If 0.484J of heat is added to 0.1372g of water, how much will the temperature increase? 0.263 °C
0.927 °c
0.843 °C
0.460 °C

ΔT = 0.843 °C

Explanation:

Given data:

Heat added = 0.484 J

Mass of water = 0.1372 g

Temperature increase = ΔT = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Specific heat capacity of water is 4.184 j/g.°C

Now we will put the values in formula.

0.484 j = 0.1372 g × 4.184 j/g.°C× ΔT

0.484 j = 0.574j/°C× ΔT

ΔT = 0.484 j / 0.574j/°C

ΔT = 0.843 °C

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2