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

1.25 Kcal

Step-by-step 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

Answer 2

`Q = 375\,cal`

Step-by-step 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\,(g)/(cm^(3))\right)\cdot (15\,cm^(3))\cdot \left(1\,(cal)/(g\cdot ^(\circ) C) \right)\cdot (25\,^(\circ)C )`

`Q = 375\,cal`