What is the specific heat of an object that weighs 15 grams and changes temperature from 30°c to 37°c when 50 j of heat is added?

1. 0.48 j/g •°c
2. 0.11 j/g •°c
3. 23.33 j/g •°c
4. 0.09 j/g •°c

0.4762 J/g°C.

Explanation:

The amount of heat released to water = Q = m.c.ΔT.

where, m is the mass of water (m = 15.0 g).

c is the specific heat capacity of water = ??? J/g°C.

ΔT is the temperature difference = (final T - initial T = 37.0°C - 30.0°C = 7.0°C).

∴ The specific heat capacity of water = c = Q/m.ΔT = (50.0 J)/(15.0 g)(7.0°C) = 0.4762 J/g°C.

the dependant variable is spilled water

the independent is the heat