You burn a 15g jellybean to warm 50 mL of water, which increases the temperature of the water 25 °C. How many calories of heat are transferred from the jellybean to the water? Assume there is no heat loss.

Explanation:

The quantity of heat transfered from the jellybean to the water is:

1.25 Kcal

Explanation:

The relationship between heat and temperature change is given by the equation:

where H = heat energy (Joules, J) , m = mass of a substance (kg) , c = specific heat (units J/kg∙K) and ΔT is the change in temperature

Given that:

mass of water (m) = density of water × volume = 1g/ml × 50ml = 50 g, ΔT = 25° C and of water= 4.18J/g/°C

Therefore H = 50g × 4.18J/g/°C × 25°C = 5225 J

The heat transferred to the Jelly bean = 5225 J = 1.25 Kcal

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.