How many grams of water when supplied with 348j of heat will gain a temperature of 5.2°c?

We can use the heat equation,

Q = mcΔT 

Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

In this problem there is no any data about initial temperature of the water. So, we can assume that given temperature of 5.2 °C as the temperature difference. 

Q = 348 J

m = ?

c = 4.186 J g⁻¹ °C⁻¹

ΔT = 5.2 °C

By applying the formula,
348 J = m x 4.186 J g⁻¹ °C⁻¹ x 5.2 °C
     m = 15.99 g

Hence, the grams of water is 15.99.

the awnser is group 12