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

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.