Question

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.

Answer 1

Answer:

$Q = 375\,cal$

Explanation:

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

$Q = \rho\cdot V \cdot c\cdot \Delta T$

$Q = \left(1\,\frac{g}{cm^{3}}\right)\cdot (15\,cm^{3})\cdot \left(1\,\frac{cal}{g\cdot ^{\circ} C} \right)\cdot (25\,^{\circ}C )$

$Q = 375\,cal$

Answer 2

Answer:

1.25 Kcal

Explanation:

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

$H=mc_p\Delta T$

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 $c_p$ 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