Question

The specific heat of ice is 0.5 calories/gram.

60 grams of ice will require
calories to raise the temperature 1c.

Answer 1

Q = 30 calories

Step-by-step explanation:

We have,

The specific heat of ice is 0.5 calories/gram.

The heat require to raise the temperature is given by :

`Q=mc\Delta T`

m = 60 grams, c = 0.5 calories/gram °C,
`\Delta T=1\ C`

So,

`Q=60* 0.5* 1\\\\Q=30\ \text{calories}`

So, 30 calories of heat is required to raise the temperature by 1 C.